^* T FOUNDATION COURSE 

IN 



Mensuration 



AND 



J 

j Mechanical Drawing I 



BY 

J. B. POORE, M. E. 

Senior Member American Society Mechanicai 
Engineers 



Foundation Course 

J. B. POORE, PUBLISHER 

Washihgton, N. J. 



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Ci3ELRIGHT DEPOSm 



FOUNDATION COURSE 

IN 

Mensuration 

AND 

Mechanical Drawing 



BY 

J. B. POORE, M. E. 

Senior Member American Society Mechanical 
Engineers 



Foundation Course 

J. B. POORE, PUBLISHER 

Washington, N. J. 






Copyright, 1916, by 
J. B. POORE 




©CLA444108 



INTRODUCTION. 

M^^HIS COURSE is fundamental. It is 
m C^ designed to help you to rise from 
^^^/ the ranks in that great army of 
men who are fighting life's battle along me- 
chanical lines. In any pursuit in life the Course 
is very helpful because it is practical. In any 
mechanical pursuit it is indispensable if you 
would be more than a drifter — a mere drudge. 
In mechanical drawing it is absolutely essential, 
for without the fundamentals taught in this 
course you cannot hope to accomplish anything. 
Thousands of young men attempt expensive cor- 
respondence courses in mechanical drawing 
every year only to drop out discouraged, be- 
cause the course assumes that they have this 
foundation already laid. They give up the 
course because they have not the basic knowledge 
on which the course is constructed. There is no 
text book that comprises more than a part of this 
basic instruction; and much that is very import- 
ant does not appear in any text book. 

In the course of a busy life along mechanical 
lines, the impression that this lack of a suitable 
text book covering these fundamentals is indeed 
very serious has been strengthened almost daily 
by the many requests for such information. Be- 
lieving that a real service may thereby be ren- 
dered to many thousands of men, the author has 
prepared this basic course in simple, practical 



4 Poore's Foundation Course 

form to help the busy man to rise to higher use- 
fulness and to more valuable service — to a bet- 
ter job and higher pay. 

TO THE STUDENT. 

There is presented to you herein a conscien- 
tious effort to help you to acquire a fundamental 
knowledge of Mensuration and Mechanical 
Drawing. The extent to which it shall be a bene- 
fit to you will be determined by your own efforts. 
Whether you learn much or little will depend 
upon the earnestness of your application and 
the fidelity with which you follow every sug- 
gestion. Do not hurry through the Course. 
Take just as much time as may be necessary to 
do the work thoroughly and with scrupulous at- 
tention to details. 

Take up the drawings in order as you are re- 
ferred to them by the text, and study them in 
connection with the text until you have a clear 
conception of the purpose of the drawing and 
of the object represented. Study the problems 
and their solution in connection with the draw- 
ing to which reference is made. Work out every 
calculation in detail, and be sure that you under- 
stand the reason for every operation before you 
perform it. Commit to memory each definition. 
Learn every rule and every principle, and study 
their application until you cannot forget them. 
When a practice is stated, make it yours. Never 
fail to look up a reference, and note well how 
it applies. Always do the work that is suggested, 
and give it your very best effort. All of this 



Pooee's Foundation Coxjbse 5 

will require time and patient endeavor; but 
that is what will make this Course valuable to 
you. Application of principles, rules and prac- 
tice is what will fix them in your mind. 

Subdivisions of sections are indicated by let- 
ters. The number references, as No. 60, No, 
83, etc., refer to the drawings. The section re- 
ferences, as Sec. 27, Sec. 24a, etc., refer to the 
textbook. 



CONTENTS, 



Drawing and Mensuration Page 7 

Sheet I, Sec. i. 

Simple Applications in Drawing Page 32 

Sheet 2, Sec. 34. 

Applications and Problems Page 39 

Sheet 3, Sec. 45. 

Applications and Problems Page 59 

Sheet 4, Sec. 61. 

Applications and Problems Page 74 

Sheet 5, Sec. 71. 

Applications and Problems Page 88 

Sheet 6, Sec. 77. 



DRAWING AND MENSURATION. 

Sec. I. The first thing in the course is the 
drawing-board (See No. i). You should have 
a board large enough to hold a 15 x i8 sheet, at 
least. It should be soft wood so as to receive 
thumb tacks readily. Have it made smooth, 
with square corners and straight and smooth 
edges, along which the head of the T-square must 
slide. Place the sheet upon the drawing-board 
and place the T-square on the sheet with the 
head of the T-square along the left end of the 
drawing-board. Make the upper or lower edge 
of the sheet coincide with the edge of the 
T-square, and fasten the sheet with a thumb tack 
in each corner. Then with the T-square draw 
a line along the upper and the lower edges of 
the sheet. These marginal lines may be half an 
inch from the edge. Then place the upper edge 
of the T-square along the middle of the sheet and 
draw a short line ^" (about) long at each end. 
These are sheet setting lines, to enable you with 
the use of the T-square to replace the sheet in 
an exactly true position at any time. Arrange 



8 Poore's Foundation Course 

the position of your drawing-board so that the 
light falls upon it from the upper left hand 
corner. Place your ink where it may not be over- 
turned. (See No. i.) 

a. No. 2 shows the T-square held in proper 
manner by the left hand while the right holds 
the ruling pen in position for drawing a horizon- 
tal line. 

b. No. 3 shows a right angled triangle placed 
in position against the upper edge of the T-square 
for drawing a vertical line. 

c. No. 4 shows a bow-pencil held in the right 
hand, the first finger over the top of the instru- 
ment, and the adjusting screw held by the thumb 
and second finger in position for changing the ad- 
justment. Two hands should never be necessary 
either to hold or to adjust a bow-pencil. 

d. No. 5 shows a bow-pencil as used in strik- 
ing a circle. 

e. No. 6 shows the compasses held in the 
right hand in position for adjusting, the second 
finger being used to open the compasses and the 
first finger to close them, while the third and 
fourth fingers grip the pivot leg. 

/. No. 7 shows the compasses in position for 
striking a circle, the head of the compasses being 
held between the thumb and first finger. 

g. Great care should be taken to obserye at 
all times your manner of holding these instru- 
ments, and to see that they are always held cor- 
rectly. Correct handling of the instruments has 
much to do with efficiency. You should fre- 



Poobe's Foundation Course 9 

quently return to sheet No. i and observe these 
several correct positions. 

Sec. 2. A point indicates position simply; it 
has not length, breadth or thickness. It may 
be indicated by a small dot. 

Sec. 3. A line has length, but not breadth or 
thickness, although in drawing we cannot repre- 
sent a line without appearing to give it breadth. 

a. A straight line, or right line, is a line ex- 
tending uniformly in the same direction. 

b. Drawing lines are heavy lines used in 
representing the outlines of an object. (See No. 
12.) Shade lines are drawing lines used to give 
an object in a drawing the appearance of stand- 
ing out in relief or else of being depressed. They 
are two to three times as heavy as an ordinary 
drawing line. Their application involves dis- 
cussions that are hardly elementary, and are not 
taken up in this course. (See No. 17.) 

c. A dimension is any measurable extent ; and 
dimension lines are light lines used to indicate 
the' limits of a dimension. (See No. 16.) 

d. A witness line is a light line showing to 
what lines a dimension extends. In practice it 
is usually used with an arrow point on each end. 
Generally the extent of the dimension is written 
along the witness line. A single witness line 
(an arrow) may be used to indicate some speci- 
fic thing. (See No. 15). 

e. Dotted lines are used to indicate the out- 
lines of any part of an object that you cannot see 
by reason of its being behind some other part. A 



10 Poore's Foundation Course 

dotted line is composed successively of heavy, 
short dashes and short spaces. (See No. 13.) 

/. A center line is drawn through centers. 
It is a light line made up successively of a long 
dash and two short dashes with short spaces be- 
tween. (See No. 11.) 

g. Broken lines are used to show the outlines 
of some object not present that you are repre- 
senting as being present with the object that you 
are drawing. A broken line is a light line made 
up successively of a long dash and a short dash 
v/ith a short space between. (See No. 14.) 

h, A horizontal line is a straight line wholly 
within the plane of the horizon; or, a straight 
line which, if extended, would touch the horizon 
at two points. (See Sec. i, a.) (See No. 2.) 

i. A vertical line is a line extending perpen- 
dicularly to the plane of the horizon ; or, extend- 
ing in the direction of the zenith ; that is, straight 
up. (Sec. I, &.) (See No. 3.) 

y. Parallel lines are straight lines lying in the 
same plane and everywhere equally distant from 
each other. The opposite sides of No. 19 are 
parallel. 

k. See No. 33. The illustration represents 
an ordinary bottle cork, and is introduced for 
the purpose of showing in simple form the use of 
the dotted line. We have a side view and a view 
of the top or larger end. Looking at the larger 
end of the cork the outline of the other end 
would not be visible; but if we could see it, we 
would see it lying within the outline of the 



Poore's Foundation Coubse 11 

larger end. We want to show it in its true posi- 
tion; therefore we show it within the outline 
of the larger end by using a dotted line. Note 
also the center lines. See also No. 60. We have 
a plain round shaft shown in a side view which 
is broken in order to economize space on the 
sheet. The set collar is to be used in connection 
with the shaft, and to show just where and how 
the collar will appear when in place, the collar 
with the set screw in it is drawn as if it were 
in place on the shaft, and the broken line is used 
to show that we are only supposing it to be 
there. Note also the horizontal center line a — b, 
and the vertical center line c — d. You have 
noticed that the end lines of the side view are 
extended, using light lines for the extensions. 
These are dimension lines. You observe also 
the light line extending from one dimension line 
to the other, and having an arrow point on each 
end. It is a witness line, showing a dimension 
of 20 ft. The side lines are parallel, and so also 
are the end lines; and both side lines and end 
lines are drawing lines, and both are straight or 
right lines. 

Sec. 4. A drawing of an object as it would 
appear from a position directly in front of one 
of its sides or ends is called an elevation. So we 
have front elevations, rear elevations, side ele- 
vations and end elevations. See No. 89. The 
two lower drawings show a front or side ele- 
vation of the bin and an end elevation. 

Sec. 5. A plan is a drawing of the top of an 
object as if viewed from a position directly 



12 Poobe's Foundation Course 

above it. See No. 89. The upper drawing is a 
plan of the top of the bin. 

Sec. 6. A section is a drawing of an object 
as it would appear to you if it were cut through 
by an intersecting plane and if the part between 
you and the cut were removed. See No. 68. It 
shows a side elevation of an I beam and also a 
section. 

Sec. 7. Often in drawing a long object, in 
order to save space on the sheet, it is represented 
as if a portion were broken out somewhere be- 
tween the ends and the remaining portions were 
brought near each other. Sometimes a cross 
section is shown between, as in No. 63. No. 60 
shows a broken elevation and a cross section at 
one end. 

Sec. 8. An inverted plan is a drawing of the 
underside of the bottom of an object, as if the 
object were turned bottom upward and were 
viewed from a point directly above it. The low- 
er drawing in No. y2 is an inverted plan. 

Sec. 9. In drawing a plan and any side or 
end elevation, the side or end to be drawn must 
be represented as if it were grasped at the bot- 
tom and swung up into the same plane as the 
top, keeping the edge that adjoins the edge of 
the top near the latter and in correct relation to 
it. 

a. Also in drawing a plan and an inverted 
plan, the inverted plan must be drawn as if the 
bottom of the object were swung around and 
upward into the plane of the top, keeping edges 



Poore's Foundation Cotjese 13 

that are on the same side of the object near each 
other and in proper relation. 

b. Also in drawing a side or end elevation 
and an inverted plan, the inverted plan must be 
drawn as if the bottom of the object were swung 
downward into the plane of the side or end, 
keeping adjoining edges near each other and 
in relation. 

c. Also in drawing elevations of sides that 
adjoin, the elevations must be drawn as if the 
sides that are being considered were swung 
around into the same plane, keeping adjoining 
edges near each other and in proper relation. 

d. Any elevation, plan or inverted plan may 
be drawn without reference to any other part 
of the drawing; but in doing so it should not be 
placed in any position that it would occupy if it 
were drawn with reference to its relation to some 
other part of the drawing. 

e. The importance of the foregoing rules 
should not be underestimated ; and in order that 
their application may be entirely clear please see 
No. 34. A is the plan of a kennel. C is the 
front elevation ; B is the left hand side elevation ; 
F is the right hand side elevation ; and D is the 
rear elevation. Each is drawn in its place and 
in its proper relation to the plan. The upper 
edge of each side adjoins an edge of the top of 
the kennel; therefore the elevations represent 
those edges as being adjacent (next to). Also 
the center lines of the plan, if extended, coincide 
with the center lines of the elevations. There- 
fore the elevations are in correct relation to the 



14 Poore's Foundation Course 

plan. C is the front elevation, and G is an ele- 
vation of the adjoining right hand side placed in 
relation to C. Adjoining edges on the kennel 
are represented as being adjacent in the drawing. 
B is the inverted plan. The edge /, of the bot- 
tom adjoins the edge / of the front C when the 
bottom is in place. In the drawing these two 
edges are represented as being adjacent, and 
their center lines extended coincide. The invert- 
ed plan B is in correct relation to the elevation 
C. B is also in correct relation to the plan A; 
because when the bottom is in place, the edge 
/ is on the same side as the edge / of the top, and 
in the drawing these edges are represented as be- 
ing nearest each other, and the center line of B 
coincides with the center line of ^. B may be 
placed also on either of the other sides of A 
and in correct relation to A, Imagine the bot- 
tom back in place ; swing its right edge down and 
over to the left and up and you will have its 
edge o adjacent to the edge o of the plan; like- 
wise if placed on the right its edge m will be ad- 
jacent to the edge m of the plan ; and if placed 
in the rear its edge n will be adjacent to the edge 
n of the plan, and in either position it will be in 
correct relation. // is a rear elevation placed 
independently (the usual practice). It is not 
placed in relation to C, and it is not in a position 
where it would be supposed to be so placed. 

Sec. ID. In drawing sections certain stand- 
ard shadings are used to indicate the material. 
(See No. 35,) a represents cast iron. (See No. 
71.) & represents wrought iron. (See No. 61.) 



Poore's Foundation Course 15 

c represents steel. (See No. 60.) d represents 
brass and copper. (See No. 62.) e represents 
lead and babbitt. (See No. 135.) / represents 
concrete. (See No. 48.) g represents wood. 
(See No. 41.) h. The lower part represents 
stone in section; the upper part represents sur- 
face of stone and glass. 

Sec. II. The art of drawing the outlines of 
objects of any shape or form on a plane sur- 
face as they appear to the eye from a given 
position is called perspective; and objects so 
represented in a drawing are said to be drawn 
in perspective. Parallel lines projected away 
from you seem to draw constantly closer to- 
gether as the eye follows them until they can 
no longer be distinguished one from another. 
(See No. 32.) The illustration represents to us 
a curb, a sidewalk, a fence, and two intervening 
grass plots. They extend side by side and par- 
allel to each other. Looking along the walk from 
any certain position, the outlines of each suc- 
ceeding section of the walk, or of each suc- 
ceeding picket on the fence, look different from 
those of the one preceding. Each successive one 
looks smaller, and slightly different in shape de- 
pending upon your position, until the eye can no 
longer distinguish one from another, and all lines 
merge into a blur which theoretically reduces to 
a point, but which practically is earlier lost to 
view. The illustration is a drawing in perspec- 
tive. No. 83 is a sketch in perspective. 

Sec. 12. The style of numerals and letters 
used on drawings is shown in Nos. 8, 9, 10 and 



16 Poobe's Foundation Course 

lo-a. Slant and spacing must be uniform. 

Sec. 13. Our standard of linear measure is the 
inch. A measure one inch (written i") in length, 
as shown in No. 18, has length only; and twelve 
of them equal one foot (written ift. or 2' — o"). 
The relation between the size of the drawing 
and the actual size of the object is called scale, 
and is usually expressed thus : i" = 12", or, J4" 
= 12"; which means that a line i" long in the 
drawing represents a dimension 12" long on the 
object, or that J4" in the drawing represents 
12" on the object. If a drawing is made to a 
certain scale, the scale should be indicated on 
the drawing. When a drawing or a part of a 
drawing is not made to scale it should be so 
indicated. No. 44 is not made to scale and is 
marked "No scale." No. 81 is drawn to scale 
J/^"=i2'^, except the height of the masonry 
foundation which is not drawn to scale; and to 
show that the dimension written is not warrant- 
ed by the scale it is marked "mk,*' which means 
that it is a "marked" dimension; that is, the 
dimension was marked on the drawing without 
reference to scale. Sketches and free hand 
drawings are made without the aid of instru- 
ments and therefore without scale. 

Sec. 14. 12 inches (12") =1 foot (i ft.) 

3 feet (3'- o") =1 yard. 

i6j4 feet =1 rod, or pole. 

25 links of 7.92 in..=i rod, or pole. 

100 links . = 1 chain. 

66 feet =1 chain. 

40 rods, or poles . .=1 furlong. 



Poore's Foundation Course 17 

lo chains =i furlong. 

8 furlongs =:! mile. 

80 chains =1 mile. 

320 rods, or poles. .=1 mile. 

5280 feet =1 mile. 

1760 yards =1 mile. 

2 yards or 6 feet . . =1 fathom. 

4 inches =1 hand. 

39.37 inches = 1 meter. 

69 % miles ...•..=! degree on 

the equator. 
1 1^72 miles ...,..= I knot. 

Sec. 15. Surface is that which has length and 
breadth without thickness. Area is quantity of 
surface. See No 19. The illustration shows 
length and breadth; and two dimensions multi- 
plied one by the other produce area. We have 
a figure i" wide and i" long, and therefore i 
square inch (written i sq. in.), which is the 
unit of measure in computing area or the amount 
of surface. 

Sec. 16. A parallelogram is any plane figure 
of four sides whose opposite sides are parallel, 
and whose angles may or may not be right angles. 
A parallelogram whose angles are right angles 
is also called a rectangle. No. 40 is a parallelo- 
gram; it is also a rectangle. The area of any 
parallelogram is found by multiplying its length 
by its breadth, or altitude (the perpendicular dis- 
tance between its parallel sides). Both dimen- 
sions should be expressed in like denominations 
before multiplying. 



18 Poore's Foundation Course 

Sec. 17. 144 sq. in =1 i sq. ft. 

9 sq. ft = I sq. yd. 



2y2}i sq. ft. 
43,560 sq. ft. 
100 sq. ft. . . 
160 sq. rds. . 



= I sq. rd. or pole. 

= I acre. 

= I square of roofing. 

= I acre. 
625 sq. links . . .= i sq, rd. or pole. 

16 sq. rds = i sq. chain. 

10 sq. chains . . . = i acre. 

640 acres zn i sq. mile. 

160 acres ......=: i quarter section. 

Sec. 18. A circle is a plane figure bounded 
by a curved line every point in which is equally 
distant from a point within, and this latter point 
is called the center. See No. 20. 

a. The curved line bounding a circle is the 
circumference, and any part of a circumference 
is an arc. As the arc d — e, 

b. A straight line passing through the center 
and terminated by the circumference is a diam- 
eter; and a straight line drawn from the center 
to the circumference at any point is a radius 
(plural radii), as r. 

c. A straight line joining the extremities of 
an arc is its chord; as d — e, 

d. A tangent is a straight line drawn from 
outside a circle so as to touch the circumference 
but not intersect it if extended; as / — g. Curved 
lines that touch but do not intersect are tangent. 

e. A segment of a circle is the portion in- 
cluded by an arc and its chord; as by the arc 
d — e and its chord d — e. 



Poobe's Foundation Coubse 19 

/. A zone is the portion between two par- 
allel chords. 

g, A sector of a circle is the portion included 
by two radii and the arc which they intercept; 
as by the radii h — d and h — e, and the arc d — e. 

h. Two or more circles are concentric when 
their centers are the same point, as in No. 
53 ; they are eccentric when their centers are not 
the same point, as in No. 49. 

Sec. 19. The degree is the unit of measure 
used in measuring the opening or difference in 
direction of two straight lines that meet or in- 
tersect each other; this opening is called an 
angle. Thus an angle is the difference of direc- 
tion of two straight lines that meet, or that di- 
verge from a common point; and this point of 
divergence is the vertex of the angle. Every 
circumference of a circle, large or small, is 
composed of 360 equal arcs; and each of these 
arcs is called a degree, and is the measure of 
the angle whose vertex is the center, and whose 
sides are the radii which include the arc. Thus, 
if from the vertex of an angle as the center you 
draw a circle, the two lines forming the angle 
become radii of the circle and include an arc; 
and this arc is the measure of the angle. An 
arc that is one-fourth of the circumference is 
an arc of ninety degrees, called a quadrant. (See 
No. 20.) It measures an angle of ninety de- 
grees. An angle that is measured by an arc 
that is half a quadrant is an angle of 45 degrees. 
An angle that is one-third of a quadrant is an 
angle of 30 degrees. An angle that is one-fourth 



20 Pooee's Foundation Course 

of a quadrant is an angle of 22y2 degrees, etc. 
(Written thus: 45 deg., 30 deg., 22 J4 deg. Also, 
a very small circle written to the right of a num- 
ber and near the top of it, is used to indicate de- 
gree or degrees. This method is employed in 
drawings. Observe its use in No. 20.) 

a, A right angle is an angle measured by a 
quadrant, or an angle of ninety degrees. 

Sec. 20. A triangle is a plane figure bounded 
by three straight lines, and having therefore 
three angles. In any triangle the sum of its 
three angles equals 180 degrees, or a half circle. 

a. In No. 20 you have illustrated three tri- 
angles usually found in a draftsman's kit. What 
angles in each? (Sec. 19-a, Sec. 20). 

ft. In any triangle the distance measured 
from the vertex of any angle perpendicularly to 
the opposite side is the altitude upon that side as 
a base. 

c. The area of a triangle is just half the area 
of a' parallelogram one of whose sides is the 
base of the triangle and the other its altitude. 
The base multiplied by the. altitude equals the 
area of the parallelogram. Therefore the area 
of a triangle is half the product of the base by 
the altitude, or the base multiplied by half the 
altitude. In No. 21. the triangle a-b-c is half 
the square a-b-c-d. Its area is half of (the 
base a-b multiplied by the altitude a-rf) . In No. 22 
the triangle a-b-c is half the rectangle a-b-d-e, 
and the area of the triangle is half (the base a-b 
multiplied by the altitude f-c). 

Sec. 21. In a right-angled triangle (See No. 



Pooee's Foundation Course 21 

21 ) the two sides that include the right angle 
are the base and the perpendicular, and the third 
side is the hypothenuse. The square of the base 
plus the square of the perpendicular equals the 
square of the hypothenuse. Therefore, to find 
the hypothenuse, add the square of the base to 
the square of the perpendicular, and extract the 
square root of the sum, 

a. To find the perpendicular, subtract the 
square of the base from the square of the hy- 
pothenuse, and extract the square root of the 
remainder. 

b. To find the base, subtract the square of 
the perpendicular from the square of the hy- 
pothenuse, and extract the square root of the 
remainder. 

c. The square of a number is the number 
used twice as a factor; that is, it is the number 
multiplied by itself. Thus in 7 x 7 = 49, 7 is 
used twice as a factor and the result is the 
square of 7; 49 is the square of 7. The square 
root of 49 is 7. Therefore the square root of a 
number is one of the two equal factors of the 
number. To find the square root of any num- 
ber observe carefully the following simple 
method : 

Separate the number into periods of two fig- 
ures each as far as practicable by commencing 
at the right hand figure. If the number con- 
tains an odd number of figures the left hand 
period will contain but one figure. If the num- 
ber is a whole number and a decimal, you should 
begin to count off the periods at the decimal 



22 Poobe's Foundation Couese 

point and count in both directions. In the case 
of decimals you always count to the right from 
the decimal point; and with whole numbers 
you count to the left. 

Then find the largest number whose square 
is contained in the left hand period. This num- 
ber is the first figure of the root, and you place 
it to the right just as you do in working long 
division. Write its square under the left hand 
period and subtract from the left hand period. 

Bring down the next period of two figures and 
write it to the right of the remainder, if there 
is a remainder; and thus form a new dividend, 
much as you do in division. Multiply the root 
already found by two and place the result to the 
left of the new dividend for a trial divisor, as in 
division. Divide the new dividend by this trial 
divisor, not including the right hand figure of 
the new dividend; that is, if the new dividend 
consists of three figures divide the trial divisor 
into the first two figures beginning at the left. 
Write the result over to the right of the first 
figure of the root as the second figure of the 
root ; also write the last figure of the root to the 
right of the trial divisor, so that the trial divisor 
with this figure added may form your true di- 
visor. Now multiply the true divisor by the last 
figure of the root and write the product under 
the new dividend. Subtract and annex the next 
period (of two figures) to the remainder for a 
new dividend. 

Double the root now found and write the re- 
sult to the left of the new dividend for a trial 



Poore's Foundation Course 23 

divisor as before; find the next figure of the 
root as above by dividing the trial divisor into 
the new dividend exclusive of the right hand 
figure ; complete the divisor as before and pro- 
ceed in the same manner until all the periods 
have been used. 

Note I. When the product of any divisor 
by the last figure of the root exceeds its corres- 
ponding dividend, the last figure of the root 
must be made less, remembering to change also 
the figure annexed to the trial divisor. Should 
a dividend not contain its corresponding trial 
divisor, a cipher must be placed in the root and 
also to the right of the trial divisor; then, after 
bringing down the next period, this last trial 
divisor must be used as the divisor of the new 
dividend. 

Note 2. When there is a remainder after ex- 
tracting the root of a number, periods of ciphers 
may be annexed as decimal iplaces, and the 
figures of the root thus obtained will be decimals. 

Note 3. If the number whose square root you 
wish to find is a whole number and a decimal, 
or if it is wholly a decimal, after separating into 
periods as above directed, you extract the root 
in the same manner as in the case of whole num- 
bers, except that if the last period on the right 
of the decimal be incomplete you should fill it 
out with a cipher. The number of decimal 
places in the root will always equal the number 
of decimal periods. 

Note 4. If the given number is a common 
fraction reduce it to its lowest terms and extract 



24 Poobe's Foundation Course 

the square root of numerator and of denomi- 
nator if both be perfect squares. If they are 
not both perfect squares, reduce the fraction to 
the form of a decimal by dividing the numerator 
by the denominator, and extract the square root 
of the decimal. If the given number is a mixed 
number, change the fraction to the form of a 
decimal before extracting the root. 

Note 5. The problems on page 25 illustrate the 
above rule. Go through each operation step by 
step in connection with the foregoing method 
and the notes to which reference is made. Then 
make up problems and extract the square root. 
To prove your work, square your root or an- 
swer; that is, multiply it by itself. The prob- 
lems that you make up will not likely be perfect 
squares, and you will not be able to find an even 
or exact root; but it will be sufficiently accurate 
if you run them out to three or four decimal 
places in the root. In the case of an uneven, 
or inexact root, of course the square of the root 
will not exactly equal the original number with 
which you started; but the difference will be 
only very slight if your work is accurate. 

Sec. 22, A polygon is any plane figure bound- 
ed by stright lines only. A regular polygon is 
one whose sides and angles are equal. Triangles 
and parallelograms are polygons, but not neces- 
sarily regular, although they may be. We have 
already considered them. To find the area of 
any regular polygon, take half the product of 
the perimeter by the perpendicular from the cen- 
ter to one of the sides. (Perimeter means boun- 



Poore's Foundation Coubse 



25 



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26 Poore's Foundation Course 

dary or contour; hence the sum of the sides). 
Or: 

Multiply the square of one of the sides by the 
number in the following table set opposite the 
number of sides in the polygon. The numbers 
in the table are constants. 

5 sides — 1.720477 9 sides— 6. 18 1824 

6 sides — 2.598076 10 sides — 7.694209 

7 sides — 3.633913 II sides — 9.365641 

8 sides — 4.828427 12 sides — 11.196152 

Sec. 23. The surface of a circle is less than 
the surface of a square whose side equals the 
diameter of the circle. (See No. 23). The 
side of the square is i". Its area is therefore i 
sq. in. To find the area of a cicrle multiply the 
square of the diam, by .7854, The diam. is i". 
The square of I'Ms i sq. in., and i sq. in. mul- 
tiplied by .7854 equals .7854 sq. in., which is 
just a little less than four-fifths of i sq. in. So 
the area of a circle of a given diameter is just 
a trifle less than four-fifths the area of a square 
W'hose side equals the diameter of the circle. 

To find the area of a circle, the circumference 
being given, square the circumference and multi- 
ply by .079577. 

a. To find the circumference of a circle, 
multiply the diameter by 3.1416. 

b. To find the diameter^ multiply the circum^ 
ference by ,3183; or, the radius being given, 
take twice the radius. 

c. The area of a sector of a circle equals the 
length of the arc multiplied by half the radius. 

d. To find the area of a segment of a circle, 



Poore's Foundation Course 



27 



find the area of the sector which has the same 
arc with the segment; also find the area of the 
triangle formed by the chord and the radii 
drawn to its extremities. Take the difference 
of these areas when the segment is less than half 
the circle, and take their sum when the segment 
is greater than half the circle. 

Sec. 24. We come now to consider bodies 
having three dimensions — length, breadth or 
width, and thickness or height. No. 24 repre- 
sents a body i'' high x i" wide x i'' long, and 
having therefore six equal sides; these sides are 
squares. Such a body we call a cube. This 
gives us our standard for measuring bodies of 
three dimensions. Thus the cubic foot is a cube 
whose dimensions are each 12" or i ft. 



12" X 



12" X 12" 



= 1728 cu. in. or i cu. ft. There- 
fore, to find the contents of cubical bodies mul- 
tiply together the three dimensions^ expressed in 
the same denomination. 



a, 1728 cu. in = i cu. f t. 

2y cu. ft zi: I cu. yd. 

128 cu. ft = 1 cord of wood. 

231 cu. in =1 I gallon. 

2150.42 cu. in = I bushel. 

2747.71 cu. in = 1 heaped bushel. 

I cu. in. wro't iron weighs 278 lb. 

cast iron weighs 26 

rolled steel weighs 283 

cast steel weighs 28 

brass weighs 31 

copper weighs 32 

bronze weighs 32 




























28 Poobe's Foundation Course 

cu. in. lead weighs 41 lb. 

" " water weighs 036 " 

cu. ft. water weighs 62.5 " 

** marble weighs 171. " 

" granite weighs 165. " 

" sand weighs 95. " 

" anthracite coal weighs. 54. " 

'* bituminous coal weighs. 50. " 

b. An ordinary brick is 2'' x 4" x 8j4". 

In a wall 8J4" thick there are 14 bricks to sq. ft. 
In a wall 12^" thick there are 21 bricks to sq. ft. 
In a wall 17" thick there are 28 bricks to sq. ft. 
In a wall 21 J^" thick there are 35 bricks to sq. ft. 
The size of bricks varies in different parts of 
the country. 

c. 16 oz =1 pound (lb.) 

100 lbs = 1 hundred weight (cwt.) 

2,000 lbs =1 ton (T.) 

2,240 lbs =1 long ton. 

60 lbs = 1 bushel of wheat. 

56 " =1 bushel of shelled corn. 

70 " =1 bushel of corn on cob. 

56 " = 1 bushel of rye. 

48 " = 1 bushel of barley. 

30 " =1 bushel of oats. 

60 " =1 bushel of clover seed. 

Sec. 25. Sawed lumber is measured by board 
measure, in which the unit is a board i ft. 
square, and i" thick, and is called i ft., board 
measure. To find the number of feet board 
measure, multiply length in feet by width in 
inches, and that product by thickness in inches, 
and divide the last product by twelve ; the result 



Poobe's Foundation Course 29 

will be the number of feet board measure. If 
the board tapers in width, take half the sum of 
the widths of the two ends for the width of the 
board. 

Sec. 26. A prism is a solid whose ends or 
bases are equal, similar and parallel polygons, 
and whose sides are parallelograms. (See No. 
25.) We have here a square prism, or one whose 
ends or bases are squares. To find the contents 
of a prism, multiply the area of one of the bases 
by the perpendicular distance between them. In 
the illustration the base is i" x i'', and its area 
is I sq. in., which multiplied by the length or 
perpendicular distance between the bases = i sq. 
in. X 13^" = ij4 cu. in., the contents of the 
prism. 

Sec. 27. A cylinder is a round body of uni- 
form diameter whose bases or ends are circular 
and parallel to each other. No. 60 represents a 
cylinder. To find the contents of a cylinder, 
multiply the area of the base or end by the alti- 
tude, or perpendicular distance between the 
bases. In No. 26 we have a cylinder whose di- 
ameter is i" and whose length is i". The area 
of one end is i x i x .7854 = .7854 sq. in. (See 
Sec. 23) and .7854 x i'' (long) = .7854 cu. in., 
the contents of the cylinder. 

a. If you imagine that you cut the lateral 
surface of a prism, or the convex surface of a 
cylinder from end to end, and spread it out on a 
flat surface, you would have a parallelogram one 
dimension of which would be the distance around 
the prism or the cylinder, and the other would 
be the altitude of the prism or the altitude of the 



30 Poore's Foundation Course 

cylinder ; and to find the area of the parallelogram 
you would multiply one dimension by the other. 
(Sec. i6). Therefore to find the lateral surface 
of a prism, multiply the perimeter of the base by 
the altitude. To find the convex surface of a 
cylinder, multiply the circumference of the base 
by the altitude. 

Sec. 28. A solid having a circle for its base 
and tapering uniformly to a point called the 
vertex is called a cone. (See No. 28). The 
slant height of a cone is a straight line drawn 
from the vertex to the circumference of the 
base. The altitude of a cone is a line drawn 
from the vertex perpendicular to the plane of 
the base. 

a. To find the convex surface of a cone, 
multiply the circumference of the base by half 
the slant height. 

b. To find the contents of a cone, multiply 
the area of the base by one-third the altitude. 

Sec. 29. A pyramid is a solid having for its 
base a polygon and for its sides triangles meet- 
ing in a common point called the vertex. (See 
No. 30). The slant height of a pyramid is a 
straight line drawn from the vertex to the mid- 
dle of one of the sides of the base. The altitude 
of a pyramid is a line drawn from the vertex 
perpendicular to the plane of the base. 

a. The lateral surface of a pyramid is found 
by multiplying the perimeter of the base by half 
the slant height. 

b. The contents of a pyramid equal the area 
of the base multiplied by one-third the altitude. 



Poore's Foundation Course 31 

Sec. 30, The frustum of any solid is the part 
that remains after cutting off the top by a plane 
parallel to the base, as the frustum of a cone 
shown in No. 29, and the frustum of a pyramid 
shown in No. 31. 

a. To find the convex surface of a frustum 
of a cone, multiply half the sum of the circum" 
ference of the two bases by the slant height, 

b. To find the lateral surface of a frustum 
of a pyramid, multiply half the sum of the per- 
imeters of the two bases by the slant height, 

c. To find the contents of a frustum of a 
cone, or of a frustum of a pyramid, multiply 
the areds of the two bases together, and extract 
the square root of the product; to this root add 
the two areas and multiply the sum by one-third 
the altitude. 

Sec. 31. A sphere is a solid bounded by a 
curved surface every part of which is equally 
distant from a point within called the center. 
(See No. 27), 

a. The surface of a sphere equals the di- 
ameter multiplied by the circumference, 

b. The contents of a sphere equctl the cube 
of the diameter multiplied by .5236, 

Sec. 32. In comparing similar surfaces, they 
are to each other as the squares of their like (or 
similar) dimensions. 

Sec. 33. In comparing similar solids, they 
are to each other as the cubes of their like di- 
mensions. 



32 Poore's Foundation Course 

SIMPLE APPLICATIONS IN DRAWING. 

Sec. 34. ''Mr. Lumberman: — Please send me 
twenty-one pieces of hemlock lumber i"x I2"x 
3'-6'^ Make a drawing showing the plan, an 
elevation of one side and an elevation of one 
end. (See No. 40). Use scale i"=i2". Make 
up a similar problem, using different dimensions, 
and make drawing to scale. (See No. 35-g-wood 
section). 

Sec. 35. ''Mr. Millman: — I want nine pieces 
of hemlock lumber delivered at the bridge that 
I am building at the Bend. They must be No. i 
in quality, 3"x I2"x 2'- 8". There must be a 
wedge-shaped piece cut off one edge of each 
joist. At a point on one end 9" from one 
edge start cutting this wedge-shaped piece, 
running out 14 inches up on the other 
other edge/' Make drawings toscale Ij4"=i2". 
Show side elevation, plan and right and left end 
elevations. A joist rests on one edge when in 
place, therefore you conceive the plan to be a 
view of the top edge, the cut having been made 
on the bottom edge. Change the size of the 
joist and make drawings to scale. (See No. 
41.) End wood is always shown as in section 
(Sec. 10). 

Sec. 36. "Mr. Foundryman: — Please cast for 
me as early as you can two corner plates, mak- 
ing the pattern at my expense and as cheap as 
possible. These castings are to be of soft iron, 
and in shape they will resemble a carpenter's 
square, with one leg 11" long and the other 7" 
long, while both are to be 3" wide and i" thick. 



Poore's Foundation Course 33 

At the end of each leg there is to be a hole 
drilled %" in diam., the center of the hole to be 
i" from the end and in the center of the width; 
also there is to be a hole drilled, same diam., in 
the corner 1J/2" from each outside edge. I en- 
close drawing.'' Make drawing to scale 3"= 12". 
(See No. 42). Show plan and end and side ele- 
vations. Make up similar problem, make draw- 
ing using different scale, and show plan and ele- 
vations of the four sides in relation to the plan. 
(See Sees. 4, 5 and 9). 



a. ''Also please make for me a cast iron, 
square, open end wrench with jaws opening 2^4" 
and 2^" deep, to fit a standard i>^" or ij4" 
square nut. Make it 16" long over all, i'' thick 
and having a handle 2" wide and rounded at the 
end. Make the head of the wrench 4%" wide 
across the jaws at the widest part, rounding the 
outside edges of the jaws with a 2^" radius 
centered ^'' from the bottom of the jaws and 
midway between them ; and then by reversing the 
curve with the same radius gracefully merge 
into the straight lines of the handle. Complete 
the curve at the end of each jaw with a J^" 
radius. Make the pattern and include the cost 
of it in your bill.'' 

b. Make a drawing of the wrench for the 
pattern maker. (No. 52). In reversing the 
curve with a 2^" radius as required, you may 
locate its center on the outside by placing the 
pivot leg of your compasses at the point in the 
curve where you wish the reverse curve to be- 
gin and describe an arc ; then place the pivot leg 



34 Poore's Foundation Course 

at the point where you wish the reverse curve 
to merge into the straight line of the handle, and 
describe an arc intersecting the other. From 
the point of intersection as a center describe the 
arc that shall be your reverse curve. Make the 
drawing half size, that is 6"=I2'^ The handle 
may be drawn full length if desired, or it may 
be broken to save space. (Sec. 7). 

c. In making castings a certain amount of 
contraction takes place in the metal as it cools, 
and this must be provided for by making the 
pattern enough larger than the casting is to be 
to allow for this contraction or shrinkage. The 
shrinkage of cast iron is about 3^" in 12" ; and in 
making a pattern for an iron casting the pattern 
maker uses a rule in which the foot is J^" 
longer than standard, and thus he provides for 
the shrinkage. The shrinkage of steel is about 
twice that of iron, and in different metals the 
shrinkage varies, so that the pattern maker must 
use different "shrink rules" in making patterns 
for the different metals. 

Sec. 37. "Mr. Machinist : — I would like you 
to make for me twenty eccentric collars of steel, 
1/4" thick, outside diameter ij^'^ and having 
an inside diameter of i", the center of which 
shall be %6 " from the center of the outside 
diameter, making the outer circle and the inner 
circle eccentric (Sec. i8-/j.). Finish the collars 
all over.'* 

a. Make the drawing with care; and make 
other drawings of eccentric circles and concen- 
tric circles, and be sure that you have well fixed 
in your mind the difference. 



Poobe's Foundation Course 35 

Sec. 38. "Mr. Contractor: — Please quote me 
your best price on 100 second quality pine frames 
as follows: The material is to be planed on 
both sides, and when assembled it is to have a 
priming coat. The stock is to be Ij4"x2", and 
is to be made up the flat way, outside dimen- 
sions of the frames to be I2"x 2'- 2". The short 
sticks are to be as long as the width of the 
frame, so they will be 12" long; and the long 
sticks are to be as long as the frame, or 2'- 2". 
Each end of each stick will be cut out for 2" by 
half the thickness of the stick across the entire 
width; or in other words, a piece will be cut 
out of each end of each stick ^" x 2" x the 
width of the stick, both cuts to be on the same 
side of the stick. This will enable a joint to be 
m.ade the same thickness as the stick." Make 
drawing showing elevation of side of frame and 
end elevation, (See No. 43), scale 3" = 12", 
(See Sec. 7 and Sec. 10, gr). Note that end wood 
is shown in these illustrations where it appears 
in elevation just as it would be shown in sec- 
tion (See No. 35). End wood is always shown 
with the sectional cross hatching as shown in 
No. 35. The metals are shown only in section 
and not in elevation (See Sec. 4 and Sec. 6). 
Change dimensions and make the drawings, us- 
ing any scale that you think will be most con- 
venient. 

Sec. 39. "Mr. Contractor: — Please make for 
me twenty-four boxes of i'^ surfaced pine lum- 
ber, securely nailed together, for use in my de- 
livery wagon. Make them 12" wide x 18" long 
X 10" deep, outside measurement. A strip of 



36 Poore's Foundation Course 

wood for a handle, ^" x 2" x 4", should be 
nailed on crosswise i'' down from the top in 
the center of the ends." Make a sketch in per- 
spective. (See No. 44). Read again Sec. 11 
and Sec. 13, the latter part. Dimensions are 
marked on a sketch without using "mk," be- 
cause being a sketch there is not supposed to be 
any reference to scale. 

Sec. 40. ''Mr. Carpenter: — There are two 
shelves much needed in our kitchen, and they 
must fit exactly in place in order to look well, 
so please make them according to the following 
exact requirements: Each shelf is to be 8J/^" 
wide X 3'- 6" long. Facing the front edge of the 
shelf, start at the left hand end and measure up 
9" along the back edge, and then cut out a piece 
6^'' long X 2j4" deep; then measure 7" more 
along the same edge and cut out another piece 
same length and depth; then, starting at the 
opposite end, on the same edge, cut out a piece 
6^" long X 25^" deep. Still facing the same 
way the left hand corner on the front edge is 
to be rounded off to a 2^'' radius. Material 
to be J^" thick, white pine, and smooth." Make 
drawing, scale i>4" = 12". (See No. 45). Write 
new specifications, changing the requirements, 
and make drawings to scale. 

Sec. 41. "Mr. Blacksmith: — Kindly make for 
me two cold chisels of good ^" hexagon tool 
steel. They should be %" wide along the cut- 
ting edge when finished and 9'' long.'' Make 
a sketch. (See No. 50). 

a. "Also please make a good poker for our 
furnace, of 5^" round iron, and four feet over 



Poore's Foundation Course 37 

all. At one end bend the iron around so as to 
form a hand hold and an oblong hole to slip the 
fingers through in grasping it. This opening 
should be i^" x 4'', and the latter dimension 
should extend in a direction at a right angle to 
the direction of the shaft; make a right angle 
bend 4" over all on the other end of the shaft, 
bending the iron in the direction of the length 
of the handle or hand hold, and flatten the end 
of the bent portion." Make a sketch of the 
poker. (See No. 46). Practice making sketches. 
Sec. 42. "Mr. Carpenter: — Please make for 
me a dog house of surfaced i" yellow pine 
boards with tongue and groove to make tight 
joints. The dimensions over all are to be — 
v/idth 24", length 3'-o", height of sides 18", 
length of roof 3'- 6", slant height of roof 2'- i" 
from eaves to apex, height to apex of roof 2'-6", 
and the two sides of the roof to form an angle 
of 90 deg. Use 3" nailing strips i'' thick 
around the inside at the bottom and lay the 
floor on them; and similarly use them around 
the inside at the top of the sides, and use one at 
the apex of the angle of the roof on the inside. 
The outside edge of the tops of the two sides 
should be chamfered off with the direction of 
the roof back to the inside edge of the nailing 
strip. In the center of one end make an open- 
ing 10" wide and 11" high above the floor, mak- 
ing the top of the opening an arc of a circle 
with a radius of 5"." (See No. 51). (Also 
see Sec. 18, a and b,, and Sec. 19). Make draw- 
ings, scale i" = 12". Show front elevation and 
side elevation in relation to it. Also draw a 



38 Poore's Foundation Course 

plan of the dog house without the roof and the 
floor. Draw a plan with roof and elevations of 
the sides and ends in relation to the plan; also 
draw them in relation to each other; draw an 
inverted plan in relation to the front elevation. 
(See No. 34, Sec. 8; also Sec. 9, a to e). Draw 
a rear elevation, or any other elevation, in an in- 
dependent position. Study this relation of ele- 
vations, plan and inverted plan until you have 
mastered it. Independent elevations are much 
used also, especially for rear mews. Study care- 
fully the use of dotted lines to show outlines 
that cannot be seen because they are behind 
something else (Sec. 3-^). Observe the proper 
use of dotted lines in drawing these several 
views. 

Sec. 43. "Mr. Foundryman: — Please make 
for me four cast steel quoits having an outside 
diam. of 5H" and an inside diam. of 3^4", 
leaving the width of the circular ring i^". 
Both edges are to be rounded on the outside as 
well as on the inside of the ring, which is to be 
3/^" thick at the outside edge and ^" thick 
at the inside. Also the ring is to rest on its 
outside edge, so that when it rests on the ground 
the top of the inside of the ring will be H" 
above the ground. On the outside edge there 
is to be a slight rounded indentation for a finger 
notch %6" deep by ^%6" across. I enclose 
drawing for the use of your pattern maker." 
(See No. 53). Make drawings. The scale is 
half size, 6" = 12". Observe the lines in the 
section to indicate material. (See No. 35). 

Sec. 44. "Mr. Cement Stone Maker: — Please 



Poore's Foundation Course 39 

make for me a cement stepping stone, or car- 
riage step, 12" high, 20^" wide and 3'-o" long. 
Make the ends round with a 10^" radius, and 
chamfer off the edge around the top down to a 
line on each side i'' from the edge/' Make 
drawing showing plan and side elevation, scale 

a. "Also please make for me a cement tie- 
post 12" square at the bottom and 8" square at 
the top, and 6'- 6" high or long. 4" down from 
the top in the center of the width of one of the 
sides, make a 2'' hole coming out in a like posi- 
tion on the opposite side. Beginning three feet 
from the bottom the edges or corners should be 
chamfered off to a depth of i'' back on each face 
and running thus uniformly to the top. Also 
in like manner chamfer off the edge around the 
top." Make elevation drawing, plan and sec- 
tional plan (Sees. 4, 5 and 6). A sectional plan 
is a view taken as though a section were made 
by a horizontal plane, and you were looking 
down at the exposed surface from a point di- 
rectly above it. Show material (concrete) in 
the section (See No. 48). Change the dimen- 
sions and make drawings to suit, scale i" = 12", 

APPLICATIONS AND PROBLEMS. 

Sec. 45. (See No. 60). We have here a 
round steel shaft 2" diam. and 20'- o" long, to 
find its contents, its weight and its entire surface. 
The set collar with set screw in it is represent- 
ed in broken line (Sec. 3, g.) because it is not 
in place on the shaft. Therefore it is repre- 
sented in broken line. It shows its relation to 



40 Poore's Foundation Course 

the shaft as it would be if in place. It has 
nothing to do with our problem. Make the 
drawing scale 6"= 12'', except in length which 
may be broken to save space. Show a side ele- 
vation and a section, indicating material in the 
latter (Sec. 10). 

a. The figure is that of a cylinder. (See 
Sec. 27). All dimensions must be considered 
in the same denomination; that is, all must be 
inches or all must be feet. To find the contents 
of a cylinder we must find the area of one end 
and multiply by the length. The end being a 
circle, to find its area we must square the diam. 
and multiply by the constant .7854, (See sec. 
2^.) which gives us 2 x 2 x .7854 == 3.1416 sq. 
in. The length is 20 ft. or 240". The area of 
one end, 3.1416 sq. in. x 240" = 753.984 cu. in., 
the contents of the shaft. 

&. The material is steel (rolled), and rolled 
steel weighs .283 lb. to the cu in. (See sec. 24, 
a.) 753.984 X .283 lb. = 213.377 lbs. There are 
16 oz. in a pound. The decimal, .377 lbs. x 16 
= 6 oz., and we have 213 lbs. 6 oz., the weight 
of the shaft. Note that the weight of iron 
in its several forms is approximately one 
quarter of a pound to the cubic inch, and that 
the total weight in pounds is approximately one 
fourth the number of cubic inches of metal. Re- 
member this. 

c. The entire surface is the area of the ends 
plus the area of its convex surface. We find 
above (a,) that the area of one end is 3.1416 
sq. in. The area of both ends is 6.2832 sq. in. 



Pooee's Foundation Course 41 

The convex surface = circumference x length. 
(See sec. 27, a.) We have the diam. and must 
find the circumference (See sec. 23, a.) Cir- 
cumference = diam. x 3.1416; 2" x 3.1416 = 
6.2832", circumference. 6.2832" x 240" 
(length) = 1507.968 sq. in., convex area. Add- 
ing to this the area of the two ends we have 
1514.25 sq. in. Reduce to square feet by divid- 
ing by 144 = 10.5 sq. ft. = 10 sq. ft. 72 sq. in., 
the entire surface of the shaft. 

d. Take a shaft of different dimensions, 
draw it to scale in broken elevation and section, 
and perform the same calculations. Do not 
pass a reference without looking it up, and al- 
ways work out calculations in detail. 

Sec. 46. (See No. 61.) The elevation shows 
the side of a bar and the section shows that its 
four sides are equal and its ends square, and 
that it is made of wro't iron. (See No. 35, b.) 
The short dimension on the elevation is not 
necessary because it is shown on the section. 
In drawing it should be left off. The same is 
true of No. 60. The bar is 4" x 4" x 20'- o" ; 
we wish to find its weight and surface. Make 
the drawing to scale 3" = 12", except in length. 
When the object drawn is long, in order to 
save space on the sheet and to save the time of 
the draftsman, it is frequently done as in No. 
61 without making a break in the elevation. You 
will notice that the drawing is not made to 
scale as regards the length. It is made any 
convenient length and the correct dimension is 
written with the letters "mk.'', which means 



42 Poore's Foundation Course 

that the dimension is marked on the drawing 
without reference to scale. Nos. 62. 65, 67 and 
68 are also examples of this practice. 

a. To find its weight we must first know its 
cubical contents. It is a square prism (See sec. 
26.) and its contents = the area of its end x 
its length. 4" x 4'^ =: area of end; therefore 
4" X 4" X 240" (length) = 3840 cu. in., the con- 
tents. A cubic inch of wro't iron weighs .278 
lb. (See sec. 24, a). 3840 x .278 lb. = 1067.5 
lbs. = 1067 lb., 8 oz., the weight of the bar. 

b. Its surface is the area of its four sides 
and its ends. Each side is 4" wide ; the sum of 
the widths of the four sides is 16" = the per- 
imeter (Sec. 2y'-ci) or "girth.'' Therefore 16" 
X 240" =: 3840 sq. in., area of four sides. The 
area of the two ends (See a,) is twice 16 sq. in. 
nr 32 sq. in., which added to the area of the sides 
=T 3872 sq. in., the entire surface of the bar. 

c. Change the dimensions of the bar and 
draw to scale. Note which are drawing lines, 
which dimension lines, which witness lines and 
which center lines, and make the proper dis- 
tinctions. 

Sec. 47. (See No. 62.) The drawing shows 
a bottom plan or inverted plan and a section of 
a half round hand-rail of brass (See No. 3Sd). 
The section shows us that the figure is that of a 
hollow cylinder cut by a plane passing longitudi- 
nally through its center, one-half being shown 
The outside diam. is 2", the length 10'- o" and 
the thickness of the wall %6" to find the weight 
of brass in the hand rail and its convex surface. 



Poobe's Foundation Course 43 

Make the drawing to scale 6" = 12", using a 
marked dimension for the length. 

a. If the drawing showed the hollow cylinder 
completed the section would show a circular ring, 
and the area of the ring multiplied by the length 
would give the contents of the wall of the hol- 
low cylinder. The area of the ring is the dif- 
ference of the areas of the two concentric cir- 
cles (Sec. 18, h,) which form the ring. The 
diam. of the outer circle is 2" ; the diam. of the 
inner circle is 2" less twice the thickness of 
the wall; 2" — (2 x^g") = 2" — ^" = i^". 
The area of a circle being the square of the 
diam. multiplied by .7854, it follows that the 
difference between the areas of two given cir- 
cles must be the difference of the squares of their 
diameters multiplied by ,785^. We have the 
diameters 2" and i^" or 1.625"; their squares 
are 4 sq. in. and 2.64 sq. in., and the difference 
of their squares multiplied by .7854 is (4 — 2.64) 
X .7854 = 1.36 X .7854 = 1.068 sq. in., the area 
of the circular ring. 1.068 sq. in. x 120" 
(length) = 128.16 cu. in. in the wall of the 
hollow cylinder. Half of 128.16 cu. in. z= 64.08 
cu. in. of brass in the hand-rail. A cubic inch 
of brass weighs .31 lb. 64.08 cu. in. of brass 
weighs 64.08 X .31 lbs. ~ 19.86 lbs., weight of 
hand-rail. (Sec. 24 a.) 

b. The convex surface of the hollow cylinder 
whose diam. is 2" = circumference x length 
(Sec. 27, a). Circumference = diam. x 3.1416 
(Sec. 23, a.). 2" X 3.1416 = 6.2832", circumfer- 
ence. 6.2832" X 120" = 753.98 sq. in., convex 
surface of hollow cylinder. Half of 753.98 sq. 



44 Poore's Foundation Course 

in. =: 377 sq. in., convex surface of hand-rail. 

Note. Half the area of the circular ring mul- 
tiplied by the length would give the contents of 
the hand-rail; and half the circumference of the 
hollow cylinder multiplied by the length would 
give the convex surface of the hand-rail. Also 
the contents of the outer cylinder less the con- 
tents of the inner cylinder = the contents of the 
hollow cylinder. Get the principles involved in 
this problem firmly fixed in your mind before 
you leave it. Make up a similar problem using 
different dimensions; make the drawing to 
scale, except the length, and work the problem. 
Observe that No. 62 presents a concave surface 
with the light falling on it from an upper left 
hand direction. The upper portion of the con- 
cave surface is shaded showing that it is in the 
shadow. No. 60 represents a convex surface, 
and for a like reason the lower portion is 
shaded. Think this over carefully. Have in 
mind clearly the difference between a concave 
surface and a convex surface and where the 
shadow would be. 

Sec. 48. No. 64 shows simply a sheet of Rus- 
sia iron, to find its weight, i sq. ft., .03" thick, 
weighing 1.21 lbs. The sheet is 24" x 72", or 
2'-o" x6'-o''and contains 12 sq. ft. 12 x 1.21 
lbs. = 14.5 lbs. = 14 lbs. 8 oz. Both views 
are broken. 

Sec. 49. No. 65 shows a steel channel of 
special design, its weight to be figured from its 
dimensions. Note that the drawing shows an 
end view and not a section, and that the ma- 
terial is not indicated on the end view. When 



Poobe's Foundation Course 45 

the material is a metal it is shown only on a 
section. Note also that in the side elevation we 
are looking at the back of the channel. The 
dotted lines show the lines of the flanges on the 
other side. One dimension is marked. 

a. The channel is composed of two flanges 
and a web (the part between the flanges), and 
both flanges and web are ^" thick. The flanges 
are 2^" over all, and the web is 6" over all, 
and between the flanges it is 5" long; so in fig- 
uring the area of the end we have three parts, 
two flanges each 2^'' and a web 5", making a 
total of 10" long by ^" wide, or 5 sq. in., the 
area of the end. The length is 20'- o" or 240". 
5 sq. in. X 240'' = 1200 cu. in. of steel in the 
channel. It is rolled steel, and not cast 
steel. Its weight would therefore be 1200 
X the weight of a cubic inch of rolled steel =: 
340 lbs. — (Sec. 24, a.) Change the dimensions 
and draw to scale showing an elevation of the 
front or channel side with an end elevation 
( End view means the same. ) Note that the flange 
ends would point toward the side elevation. 
(See sec. 9, c and e.) Be sure that this is clear 
to you before you leave it. Use a marked di- 
mension for the length and show a section at 
the right. 

Sec. 50. (See No. 67.) In the drawing we 
have represented the back view of a special an- 
gle beam of rolled steel and an end view. Figure 
the weight from the dimensions. To find the 
area of the end, the web is 3>4" long and the 
flange is ij^" over all, and it projects ij^" 
less ^'^ or ij^". We have a total length of 



46 Poore's Foundation Course 

3J4'' added to i>^", or 4^" x ^" wide = 1.64 
sq. in., the area of the end, which multiplied by 
the length = 1.64 sq. in. x 216" = 354.24 cu. in. 
of steel in the beam. 354.24 x .283 lb. = 100 
lbs. 4 oz., the weight of the beam. Draw a 
front view of the beam with a right hand end 
view. The flange would point the same way 
that it does in No. 67. 

Sec. 51. No. 66 shows a broken side view and 
an end view of a wro't iron tube. Note that the 
material is indicated on the section but not on 
the end view. The inside and the outside diam- 
eters and the length are shown. It is ^required 
to find the weight of the tube. The thread 
shown will be discussed farther on in the 
course. 

a. The figure is that of a hollow cylinder, 
and we will find the area of the circular ring 
shown by the end view. Its outside diam. is 
8.625" and its inside diam. is 7.982". The dif- 
ference of their squares x .7854 = area of cir- 
cular ring. (See sec. 47-a.) The square of 
8.625" = 74.3906 sq. in. The square of 7.982" 
==: 63.7123 sq. in. The difference is 10.6783 
sq. in., and 10.6783 sq. in. x .7854 = 8.3867 sq. 
in. — the area of the circular ring, or of the end 
of the tube. The area of the end x the length 
=:: the contents. 8.3867 sq. in. x 120" =: 1006.404 
cu. in. of wro't iron in the tube, weighing .278 
lbs. to the cu. in. 1006.404 x .278 lbs. =: 279.78 
lbs. = 279 lbs. I2j4 oz., the weight of the tube. 

b. Assign new dimensions to the tube, make 
the drawings to any convenient scale, and find 



Poore's Foundation Course 47 

the weight of the tube. Observe to hold your 
instruments correctly in drawing. 

Sec. 52. (See No. 69.) The drawing repre- 
sents a cast steel roller comprising three parts, 
two cylinders and a square prism, to find its 
weight. The diam. of the smaller cylinder is 5", 
and 5" X 5'' X .7854 = 19.63 sq. in., the area of 
the end of the smaller cylinder. 19.63 sq. in. x 
2>4" (length) = 49 cu. in., the contents. The 
diam. of the larger cylinder is 6", and in like 
manner 6" x 6" x .7854 x 3^^" = 99 cu. in., 
the contents of the larger cylinder. The sguare 
prism is 2^" square by 2'' long; its contents 
must be 2j4'' x 2j^" x 2" = 12.5 cu in. The 
total contents of the roller are the sum of 49 
cu. in., 99 cu. in. and 12.5 cu. in., or 160.5 cu. in. 
Cast steel weighs .28 lb. to the cubic inch (Sec. 
24, a.). 160.5 X .28 lb. = 44.9 lbs., the entire 
w^eight. Note the left hand end view, and the 
right hand end view showing only half the view 
on account of lack of space, and observe the use 
of dotted lines to represent the outlines that 
could not be seen from the view point. Can 
you explain the differences of those end views 
or elevations? Assign new dimensions to the 
roller and make a drawing to scale. Show a 
horizontal section. The drawing will be a sec- 
tional plan. The outlines will be just the same 
as those of the side elevation in No. 69, except 
for the changed dimensions. 

Sec. 53. No. 63 shows a broken elevation 
and a section of a wro't iron bridge link i'' 
thick, 6" wide and 4^- 6" long over all, having 
the ends rounded off to a 3'' radius and a 2" 






48 Poore's Foundation Coubse 

hole drilled in each end in the center of the 
width and 3" from the end. What is the weight 
of the bridge link? Make the drawing to scale. 

The link being just 6" wide a 3" radius will 
describe an arc of 180 degrees, or a half circle 
(Sec. 19.) in rounding the ends, and the por- 
tion of the link between these half circles will 
have the form of a prism whose dimensions 
are i" x 6" x 4'-o", or i" x 6" x 48" = 288 
cu. in. in that portion of the link. The half 
circles on the ends having each a radius of 3 
are equal to a circle having a diameter of 6 
which forms the end of a cylinder i" long, 
whose contents = 6" x 6" x .7854 x i" = 
28.27 cu. in., contents of the ends. Add this to 
288 cu. in., and the total is 316.27 cu. in. Now 
deduct the holes. They are 2" in diam. and i" 
deep or long, so their form is that of a cylinder 
(Sec. 2^.), and the contents of each one is 2" x 
2" x .7854 X \" = 3.14 cu. in. 2 X 3.14 cu. in. 
= 6.28 cu. in., contents of two holes. Subtract 
6.28 cu. in. from 316.27 cu. in. = 310 cu. in., 
contents of link. 310 x .278 lb. = 86.18 lbs. = 
86 lbs. 3 oz., the weight of the link. Make up a 
similar problem, work it out and make drawing. 

Sec. 54. (See No. 68.) We have an eleva- 
tion of a side and a section (taken at A A) of 
a steel I beam (No. 35, c.) that is 6>4" high, 7" 
wide and whose flanges and web are each i" 
thick, and which is 30'- o" long. We wish to 
find its weight. The rule for finding the con- 
tents of prisms (Sec. 26.) and cylinders (Sec. 
2y.^ applies to bodies of any shape whose ends 
are equal, similar and parallel, and a cross sec- 



Poobe's Foundation Course 49 

tion of which at any point would be equal and 
similar to the ends; that is, the area of the end 
multiplied by the length equals the cubical con- 
tents. See how this applies to the preceding 
numbers on Sheet No. 3, and get the principle 
firmly fixed in your mind. 

a. The top and bottom flanges are each i" 
X 7", and the web, or the part that connects the 
flanges, is i" thick x 6^2" long less twice the 
thickness; that is, it is i" x 4JE^". Putting these 
together, the two flanges and the web are 7" 
and 7" and 4>^" = i8>4" long by i'' wide = 
18^ sq. in., area of the end of the I-beam. i8j4 
sq. in. X 360" (30'- o") = 6660 cu. in. of steel, 
weighing .283 lb. to the cubic inch = 1885 lbs. 
Would an end view have the shading lines to in- 
dicate material? (Sec. 49.) Change the di- 
mensions, making the beam larger and draw it 
to scale or to a smaller scale, and find its weight. 

Sec. 55. No. 70 shows a plan and a side ele- 
vation of a cast iron tamping head having a 
square base and a hollow cylinder with a hole 
through its sides to receive and fasten the handle. 
Figure its weight from the drawing. First make 
the drawing one quarter size; that is, 3"= 12 
The base is a rectangular prism, 6'' square by 2 
high, and contains 6" x 6" x 2" =z 72 cu. in. 
Then the cylindrical part, regarding it as being 
solid, is 3'' in diam. and 3" high. (Sec. 27.) Its 
contents must be 3" x 3" x .7854 x 3" long, or 
high = 21.21 cu. in., which added to the base 
makes 93.21 cu. in. Now deduct the 2" hole 
for the handle and the %" core hole for fasten- 
ing the handle. The 2" hole is 3^^" deep. Its 






50 Poore's Foundation Course 

shape is that of a cylinder, so its contents must 
be 2" X 2" X .7854 X y/2'' = II cu. in. The 
%" hole penetrates the wall of the hollow cyl- 
inder in two places. This wall is J^'' thick 
(y2 of 3"— 2''), and the hole through it is %" 
diam. by a depth or length of J4''. Its ends are 
equal, similar and parallel (although curved), 
and it is a cylinder. Its contents must be %'' x 
%" X .7854 X >^'' = .3 cu. in.; and two such 
holes contain .6 cu. in., which added to the con- 
tents of the 2" hole makes 11.6 cu. in. De- 
ducting 1 1.6 cu. in. from 93.21 cu. in., we have 
81.61 cu. in. of cast iron weighing .26 lb. to the 
cubic inch, or 21 lbs. 3.5 oz., the weight of the 
tamping head. 

a. Increase the dimensions, draw to any 
scale and figure the weight. Explain the use of 
dotted lines in the drawing. 

Sec. 56. (See No. 71.) We have two views 
of a cast iron pole link and a section of the 
middle part, which connects the two hollow cyl- 
inders, taken at C — C as indicated on the draw- 
ing. You will notice that where the middle por- 
tion connects with the hollow cylinders the corn- 
ers are filled in so as to make a curve instead of 
an angle in the outline. You will notice the same 
thing in No. y2 where the middle portion con- 
nects with the bases. It is a rounding of in- 
side corners by filling in to a curved line, and 
this filling in of a corner is called a fillet. For 
the purpose of this course we will make allow- 
ance for the fillet approximately, as is usually 
done in practice. We wish to figure the weight 
of this pole link. 



Poore's Foundation Course 51 

a. The two hollow cylinders on the ends are 
each 6" diam. outside and 4" inside. (See sec. 
47, a.) The area of the circular ring is the dif- 
ference of the areas of these two circles, and the 
difference of their areas is the difference of the 
squares of their diameters multiplied by .7854. 
Therefore (6 x 6) — (4 x 4) x .7854 = the area 
of the circular ring =: 15.7 sq. in. 15.7 sq. in. 
X 3" (length of hollow cylinder) = 47.1 cu. in., 
and this multiplied by 2 equals 94.2 cu. in., the 
contents of the two hollow cylinders. Now take 
the middle part which is 2)^" x 3" x its length. 
We find that the length from one hollow cyl- 
inder to the other along the center line is 4" 
(by deducting the radii of the two outer circles 
from the dimension from center to center.). 
Keeping in mind the scale of the drawing as we 
study it, an allowance of ^'' to be added to the 
length of this middle portion seems to be suffi- 
cient to cover the material in the fillets, so we 
will consider the length of the middle portion 
along its center line to be 4j^". 2^" x 3" x 
4 14" = 31.9 cu. in. Add 94.2 cu. in. and we 
have 1 26. 1 cu. in., the total contents, weighing 
.26 lb. to the cubic inch, or 32 lbs. 12^ oz., the 
weight of the pole link. Change the dimensions 
and draw to scale and work out the problem. 

Sec. 57. No. 72 shows a side elevation and 
an inverted plan (Sec. 8.) of a cast iron pedestal 
having one round base and one square base con- 
nected by a cylindrical portion, through one 
side of which there is a slotted hole communi- 
cating with a round hole that extends through 
the center of the pedestal along its longitudinal 



52 Pooee's Foundation Course 

axis. The inverted plan shows whether the 
round base is at the bottom or at the top. Where 
is it? How can you tell? Make the drawing to 
scale 3" = 12". Find the weight of the pedestal, 
a. We have to figure the contents of the 
round base, the square base and the cylinder, 
and deduct the contents of the slotted hole and 
the round hole. The square base is at the top. 
It is 7J4" square by i" thick, and its contents = 
7^4" X 7J/^" X I" = 56.25 cu. in. The round 
base is at the bottom, and its contents =: area 
of end X length or thickness. The area of the 
end or face of the round base is 7^" x 7j4" x 
.7854 = 44.18 sq. in., and this multiplied by the 
thickness = 44.18 cu. in., contents of round 
base. The middle portion has the form of a 
cylinder. (Sec. 27.) Its contents = area of 
base X length. Its diam. is 4" and its length is 
9^", (Find these dimensions from those 
given), and 4" x 4" x .7854 x 9^'' = 122.52 cu. 
in., its contents. Adding to this the two bases 
we have 222.95 cu. in., total contents. Now the 
round hole through the center of the pedestal is 
IT^" long by 2^" diam. Its contents must be 
2>^" X 2j^" X .7854 X ii%" = 57.68 cu. in., 
contents of round role. The slotted hole is ij4" 
wide X 7 ^" long x thickness of the wall 
through which it is cut. From the dimensions 
given we find that this wall is half the differ- 
ence of the diameters of the round hole and 
the outer cylinder, or ^". Therefore the con- 
tents of the slotted hole = H'' x ij^" x 7^" 
— 8.72 cu in. Adding 57.68 cu. in. we have 
66.4 cu. in., which we subtract from the gross 



Poobe's Foundation Course 53 

contents, 222.95 cu. in., and have 156.55 cu. in. 
of cast iron in the pedestal, or 40 lbs. 1 1 oz. 

b. Draw a plan of the top of the pedestal 
to scale. Also draw an elevation of the side 
opposite the side that is shown, or the rear. 
Draw a right hand elevation. Be careful to use 
dotted lines correctly, to show outlines that can- 
not be seen from your point of observation in 
drawing any elevation or plan. Draw a section 
and indicate on the drawing where it is taken. 
(See No. 71.) 

Sec. 58. No. 74 shows two views of a cast 
steel lever, to find its weight, approximating ex- 
actness closely enough for practical purposes. 
You will see that we have a hollow cylinder and 
a bar of uniform thickness but tapering in width. 
We will deduct the key way shown in the hollow 
cylinder, but we will disregard the fillet where 
the latter joins the bar, and we will also dis- 
regard the rounding of the edges of the bar, al- 
lowing this to offset the fillet. The hollow cyl- 
inder is 2^" diam. outside by i>4" diam. in- 
side by 2" long. (See 47,^.) Its contents must 
be (2j4" squared) — (i>4'' squared) x .7854 x 
2" (the length), or 6.25 sq. in. — 2.25 sq. in. x 
.7854 X 2" = 6.28 cu. in. The length of the 
bar or handle along its center line to the hol- 
low cylinder is 2y'^ less the radius of the out- 
side diam. of the hollow cylinder, or 25^" long. 
At the wide end it is 2j/^" wide and at the other 
end it is 1^4" wide. The average width is half 
the sum of the end widths, or 23/^", average 
width. So the handle is %" (thick) x 2>^" x 
25^" zn 41 cu. in., contents of handle. Adding 



64 Poore's Foundation Coubse 

6.28 cu. in. we have a total of 47.28 cu. in. The 
key way in the hollow cylinder is J4'^ x J^'' x 2" 
= % cu. in., or .25 cu. in. Subtracting from 
47.28 cu. in. we have 47.03 cu. in. of cast steel 
weighing .28 lb. per cubic inch, or 13.17 lbs. = 
13 lbs., practically, because of the approxima- 
tions. 

a. Notice that the section is not drawn ex- 
actly to scale and is so marked. Increase the 
dimensions and draw to scale, and figure the 
weight. 

Sec. 59. (See No. 73.) The drawing shows 
a cast iron cleaning door, having a depressed 
panel containing two bosses on its front and a 
panel in relief, or raised panel, on its back. 
The drawing comprises a front elevation, a right 
hand elevation, a vertical section taken along 
the vertical center line, and a sectional plan 
bisecting one of the bosses. Study the drawing 
and the dimensions very carefully until you have 
in your mind a clear conception of just what the 
draftsman means to show; then find the weight 
of the door. We will regard the door as being 
i" thick X 8" wide x 10" long or high; we 
will add the raised panel on the back, deduct the 
depressed panel on the front, add the two bosses 
and deduct the core holes. The contents of the 
door 1=: i" X 8" X 10" = 80 cu. in. The raised 
panel on the back is J4'' x 7" x 9" and con- 
tains 15.75 cu. in., which added to the door =: 
95.75 cu. in. The depressed panel on the front 
contains J^" x 5" x 8" =: 20 cu. in. Deducting 
20 cu. in. we have 75.75 cu. in. The two bosses 
are each 2" in diam. by J^" high. Regarding 



Poore's Foundation Course 55 

them as solid each contains 2" x 2" x .7854 x 
y2^ = 1.57 cu. in. And both contain 2 x 1.57 
cu. in. 1= 3.14 cu. in. Adding this we have 
78.89 cu. in. Now deduct the core holes (two). 
They are y%^^ diam. by i^" long or deep. Each 
contains %" x J^'' x .7854 x i^" = .75 cu. in., 
and in both there are 1.5 cu. in. Deducting this 
there is left 77.39 cu. in. of cast iron in the door, 
weighing 77.39 x .26 lb. = 20 lbs. 2 oz. 

a. Make the drawing to scale, and add an 
inverted plan of the lower edge of the door. 
Make an elevation of the back of the door and 
place it independently. Double the dimensions 
and make drawing to same scale and figure the 
weight. Core holes are made when the metal is 
cast. They are not drilled. 

Sec. 60. Nos. 75, 76, "JTy 78 and 79 are part 
of the detailed drawings that go with the roof 
truss on Sheet No. 5, as they represent parts of 
that construction. They are introduced here to 
further illustrate methods and processes in shop 
arithmetic. 

a. We will now take up No. 75. It shows 
a cast iron washer, being 3" in diam., %" thick 
and having a core hole J^" in diam., so that 
it will easily slip over a %" bolt. It is called 
a ^'' washer. We will figure the material in 
it and its weight. You will at once recognize 
that in form it is a hollow cylinder, and the 
square of 3" minus the square of %" x .7854 x 
%" nz the contents; or, (9 — 4:%^ ) ^ .7854 x 
%" = 5.658 cu. in., weighing .26 lb. to the cu. 
in, or a total weight of i lb. yy2 ozs. 

&. No. 76 shows a cast iron bevel washer. 



56 Pooee's PouNDATio]sr Course 

also ^" size. It is 3" in diam. and has a %" 
core hole. It is beveled off from a thickness of 
i^" down to %^\ We must figure it in two 
parts. First figure the hollow cylinder whose 
diameters are 3" and J^" and whose length is 
J4"; then figure half of a hollow cylinder whose 
diameters are the same and whose length is ij4". 
The contents of the former are (3" squared — 
%" squared) x .7854 x ^4" = 1-617 cu. in. 
The contents of the latter are J4 of (3" squared 
— %" squared) x .7854 x i>4" = 4.85 cu. in., 
which added to 1.617 cu. in. =: 6.467 cu. in. 
This multiplied on the weight of a cu. in. of cast 
iron =: I lb. II ozs., the weight of the washer. 

c. No. yy shows a i^' cast iron washer hav- 
ing a diam. of 4", a core hole of i^" and a 
thickness of i". Figure its weight. The method 
has already been illustrated a number of times. 
See sec. 60-a, 

d. No. 78 shows a peak washer to be used 
on an inch bolt at the top or peak of a roof 
truss having a fifty per cent, rise, or a rise of 
6" in 12" or 3" in 6". We have two views, a 
plan and a side elevation. The dimensions be- 
ing shown, we desire to find its weight. We 
must figure it in three parts — the middle part 
and the two sides or wings. We must make a 
drawing so that from it a pattern for a casting 
m.ay be made that will fit to the rafters of the 
truss; and the angle or direction of the sides 
must conform to the direction of the rafters. 
In drawing the view of the side of the washer 
as you have it in your mind, draw the lines a — 
b and c — d representing the upper and lower 



Poobe's Foundation Coubse 57 

edges of the middle part of the washer. Then 
frorri the point d draw d — h perpendicular to 
c — d and representing a length of 3". From h 
draw h — i perpendicular to d — h and represent- 
ing a length of 6". The points d and i give you 
the direction for drawing the line d — e so that 
the washer will fit the truss. You can then com- 
plete the wing d-e-f-b. Draw the other wing in 
the same manner. 

To find the contents of the washer we must 
find the area of the side as shown and multiply 
by the width; then deduct the core hole. The 
area of the middle part a-b-c-d is ij4" x 4" = 
5 sq. in. The two wings are alike, so we will 
consider b-d-e-f. It is not a parallelogram, but 
the sides b — d and e — f are parallel, and its 
area is half the sum of its parallel sides multi- 
plied by the perpendicular distance between 
them; b — d is i>^" and e — f is %"; the per- 
pendicular between them is 3". ^ of (i^" plus 
%") X 3" = 3 %6 sq. in., the area of side of 
one wing ; area of two wings 1= 6}i sq. in. Add- 
ing 5 sq. in. and 6}i sq. in. we have 11^ sq. in., 
the area of the side of the washer; and multi- 
plying by 6'^ the width, we have 68j4 cu. in., 
from which we must deduct the hole. The hole 
is ij^" in diam. and ij^" deep or long. In 
shape it is a cylinder. Its contents must be the 
square of i>^" x .7854 x ij4" = ij4 cu. in. 
(1.25 cu. in.) Deducting 1.25 cu. in from 68>4 
cu. in. (68.25 cu. in.) we have 67 cu. in. of cast 
iron weighing .26 lb. to the cu. in., or a total 
weight of 17.4 lbs. 

Be sure that you understand the method used 



68 Poore's Foundation Course 

to find the angle of the lines representing the 
under side of the wings to make them fit rafters 
having a rise of 3" in 6'^ Draw another peak 
washer to fit on rafters having a rise of 4" in 
12"; and another for rafters having a rise of 8" 
in 12". Draw them to any convenient scale and 
find their weight. 

e. No. 79 shows a plan, a side view and an 
end view of a toe plate for the roof truss shown 
on Sheet No. 5. It is the plate on which the 
lower end of the rafter rests, made so that the 
rafter cannot slip. The plate is i" x 6" x 12", 
and having three projections or ribs running 
across its width, two on one side and one on the 
other. We wish to know how much cast iron is 
in it. The body of the plate being i" x 6" x 12" 
contains J2 cu. in. Two of the ribs are i'' wide 
X i" high X 6" long, and each contains therefore 
6 cu. in., and two of them contain 12 cu. in. The 
third rib is i" high on one side and J^'' high 
on the other and i" wide. The area of its end 
multiplied by its length = its contents. The sum 
of its parallel sides is i" plus J4", and half their 
sum is %" which multiplied by i", the perpen- 
dicular distance between them, = ^ sq. in., 
the area of the end; multiplying by the length, 
6", we have 4^^ cu. in., the contents of the third 
rib. Adding J2 cu. in. and 12 cu. in. and 4^ 
cu. in., we have 88^ cu. in., the contents of the 
toe plate, less the hole. The hole is J^" wide 
and 3" long, having rounded ends. The two 
rounded ends are each a half circle and if put 
together Mrould make a circle J^" in diam. 
whose area is the square of %" x .7854 = .6 



Poore's Foundation Course 69 

sq. in. The parallelogram between these semi- 
circular ends is J^" x 2j^" and contains 1.86 
sq. in. Adding the ends we have 2.46 sq. in., 
the area of the hole. Multiply this by the depth 
of the hole or i" and we have 2.46 cu. in., the 
contents of the hole. Subtract this from 88J/2 
cu. in. (88.5 cu. in.) and we have 86.04 c^- i^-> 
weighing .26 lb. to the cu. in., or a total weight 
of 22 lbs. 6 oz. In the drawing the hole is not 
made to scale, and is so indicated. In drawing 
it to scale the ends should be rounded to a 
%6" radius, making the diam. %". 

Sec. 61. No. 82 represents a pile of coal 
against a side wall in a cellar up to the height 
of the bottom of the window frames from one 
end wall to the other and extending out from 
the base of the side wall uniformly a distance 
of 12'- o'^. How many tons of anthracite coal 
are in the pile? The drawing shows an inside 
elevation of the side wall of the cellar and the 
face of the coal pile, and a section of both. The 
scale is ^" == 12". Make the drawing first. 
We find from the drawing that the coal pile is 
in the form of a triangular prism (Sec. 26) ; 
its bases, or ends, are equal and parallel tri- 
angles whose altitude is 7'- 6", and whose base 
is 12'- o". The area of an end of the coal pile 
is therefore J^ of (12'- o" x 7j^') = 45 sq. ft. 
(Sec. 20, r.), and 45 sq. ft. x 16'- o" = 720 cu. 
ft. of coal. A cubic foot of anthracite coal 
(Sec. 24,a.) weighs 54 lbs. Therefore 720 x 
54 lbs. =: 38,880 lbs. = 19T. 880 lbs. of coal in 
the pile. 

a. How many square feet of surface in the 



60 Poobe's Foundation Coubse 

face of the pile ? This surface is a parallelogram 
whose length is i6'- o", and whose width is the 
slant height of the pile, or the hypothenuse of 
the right-angled triangle which forms the out- 
line of the end of the pile or of a section. The 
hypothenuse (Sec. 21.) of a right-angled triangle 
nr the square root of the sum of the squares of 
the other two sides. The base of the triangle 
is 12'- o", and the altitude is 7^4'; the squares 
of these numbers are 144 sq. ft. and 56.25 sq. 
ft.; and their sum is 200.25 sq. ft. The square 
root of 200.25 sq. ft. is 14.15 ft., the width of 
the face of the coal pile. The width x the 
length = 14.15 ft. X 16 ft. = 226.4 sq. ft., 
the surface of the face of the coal pile. 

b. Note the stone walls shown in section, 
and also the coal pile. (See No. 35-A.) Also 
note how the glass in the cellar windows is 
shown, and also the face of stone in walls not 
shown in section, as indicated in the upper part 
of No. 35-/^. (See also Nos. 81, 83, 84 and 86.) 
Give the problem new dimensions and make a 
new drawing and new calculations, putting 
great care on the details of the drawing. 

Sec. 62. In No. 81 we have an elevation and 
the plan of a concrete cone on a stone founda- 
tion. Find how many cu. yds. of masonry are 
in the foundation; how many cu. ft. of concrete 
in the cone; and the cost of painting the cone 
at y2 cent a sq. ft. The base is in the form of 
a cylinder (Sec. 27.) whose diam. is 4'- 8" and 
whose length is 3'- 6". The area of the end of 
the cylinder (Sec. 23.) is 4^' x 4%' x .7854 
=r 17.1 sq. ft., which multiplied by 3J4' (length 



Poore's Foundation Course 61 

or height) = 59.8 cu. ft, or 2.18 cu. yds. of 
masonry in the foundation. 

a. Now we consider the upper portion, or 
the cone (Sec. 28.) The area of the base of 
the cone is 17. i sq. ft. as we found above, since 
the base of the cone coincides with the end of 
the cylinder which forms the foundation; 17. i 
sq. ft. X (Ys of 4'-o") = 17.1 sq. ft. x % ' = 
22.8 cu. ft. of concrete in the cone. (Sec, 28,&.) 
The painted surface will be the convex surface 
only, and the convex surface equals circum. of 
base X J4 of slant height (Sec. 28, a.). The cir- 
cumference equals diameter x 3.1416 (Sec. 
23, a.). 4^' X 3.1416 = 14.66', the circumfer- 
ence; and this multiplied by (J4 of 4'- 7") (or 
14.66' X ^%40 = 33.6 sq. ft. of surface to 
paint at J4 cent a sq. ft. =: 17 cents, cost of 
painting. Make drawing to scale, as given; 
then change dimensions, make drawing to scale 
and work problem. Study references and learn 
rules. 

Sec. 63. No. 80 shows a small concrete pier 
whose ends are parallel squares and whose sides 
are slanting. Its form is that of a frustum (Sec. 
30.) of a pyramid (Sec. 29.) We have shown 
a side elevation and the plan, with dimensions, 
to find the cubic feet of concrete in it, and the 
cost of painting the exposed surface (standing 
on its larger base) at ^2 cent a sq. ft. Make the 
drawing first. 

a. We are to find the contents of the frus- 
tum (See Sec. 30,c). The lower base is 18" 
square, and the upper base is 12" square. Their 



62 Poore's Foundation Course 

areas (Sees. 15 and 16) are respectively 324 sq. 
in. and 144 sq. in., and the product of their areas 
is 46,656. Extracting the square root, we have 
216, to which we add 324 and 144 = 684 sq. in. 
Alultiply by Yz of 22" (or 684 sq. in. x7>^'0 
=1 5016 cu. in. zn 2.9 cu. ft. of concrete in the 
pier. (For square root see Sec. 21-c.) 

h. To find the surface of the sides we must 
know the slant height, which would be the short- 
est distance between the edges of the bases 
measured at the middle of a side. In this case 
the sides being equal, the slant height of the 
sides is the same. If you could drop a perpen- 
dicular line from the edge of the upper base it 
would touch the lower base 3" from its edge, and 
you would have a right angled triangle whose base 
is 3", perpendicular 22^^ and whose hypothenuse 
is the slant height of the frustum. We must find 
the hypothenuse (Sec. 21.) The square of 3" 
added to the square of 22^^ = 493 sq. in., and 
the square root of 493 sq. in. = 22.2", the slant 
height. The area of each side would be half 
the sum of its parallel sides or ends multiplied 
by the slant height or shortest distance between 
them ; so the area of the four sides must be half 
the sum of the perimeters of the bases multi- 
plied by the slant height. (Perimeter means 
the distance around, or the sum of the four 
sides.) The sum of the perimeters of the bases 
is 120", and half of this, or 60" x 22.2" = 1332 
sq. in., the area of the four sides; add the area 
of the top or upper base, and we have 1476 sq. 
in. =r 10.25 sq. ft. of surface to be painted, at 
Yz cent per sq. ft. = 5 cents, cost of painting. 



Poore's Foundation Course 68 

Make a similar problem by changing dimensions ; 
make the drawing and work the problem. 

c. Draw a figure having two round bases in- 
stead of square ones. It will be a frustum of 
a cone. Add the dimensions according to scale, 
and find the contents and convex surface. The 
principles are the same as in above problem. 

Sec. 64. In No. 89 we have the plan, an end 
elevation and a side elevation of a chest of 
bins, in each of which are four upright corner 
pieces for nailing strips, made by ripping 
diagonally 3" square sticks. A is filled with 
shelled corn, B with wheat and C with oats. 
What weight of grain is in the chest? How 
many square inches of tin will it require to line 
the sides and bottom of the chest on the inside? 
Make the drawing to scale with great care. 

a. We must first find the contents of each 
bin without considering the corner pieces, and 
then deduct the corner pieces. A is 3'-o" x 
4'-o" X 3'-o" and contains 36 cu. ft., or 62,208 
cu. in. The corner pieces are triangular prisms 
(Sec. 26.) The area of an end of a corner 
piece (Sec. 20,c) is half of 3" (base) x 3" 
(altitude) = ^ x 3" x 3'' == 4>4 sq. in., and 
multiplying by the length or 36" gives us 162 
cu. in. in each corner piece. There are four of 
them in a bin, or 648 cu. in. of corner pieces in 
each bin. 62^208 cu. in. less 648 cu. in. = 61,560 
cu. in. In each bin there is a batten or cleat 
on the under side of the lid which must be de- 
ducted. The batten is i>^" x 8" x 3'- i", and 
contains 370 cu. in. 61,560 cu. in. less 370 cu. 
in. =r 61,190 cu. in. of shelled corn in A, or 



64 Poore's Foundation Course 

28.45 ^^' of shelled corn weighing 1593.2 lbs. 
(Sec. 24-a-r.) B is the same size and contains 
61,190 cu. in., or 28.45 ^^' of wheat, weighing 
1707 lbs. C is 2'-o" X 3'-o" X 4'-o" and con- 
tains 41,472 cu. in. less the corner pieces and the 
batten. The corner pieces contain 648 cu. in. 
and the batten contains 370 cu. in., as found 
above. Together they contain 1018 cu. in. 
41,472 cu. in. less 1018 cu. in. =z 40,454 cu. in., 
or 18.8 bu. of oats, weighing 564 lbs. in C And 
in all three bins there is a total of 3864 lbs. of 
grain. 

b. Now we want to know how many square 
inches of surface there are to be lined with 
tin. The sides, the end and the bottom of the 
chest are to be covered, and we assume that it 
is done before the partitions are put in. The 
four corner pieces of the chest must be in place 
in order to nail the chest together, and they will 
therefore be in place when the lining is put in. 
The end walls are each 42" long between corner 
pieces and 36" high, and contain 15 12 sq. in. 
In the two end walls there are 3024 sq. in. The 
sides are 92" long between corner pieces and 
36" high, and contain 3312 sq. m. each. In the 
two sides there are 6624 sq. in. The bottom is 
98" long and 48" wide and contains 4704 sq. in., 
from which we must deduct the ends of the four 
corner pieces. We have found (a) that the 
area of the end of a corner piece is 4.5 sq. in., 
and four corner pieces would cover 18 sq. in. of 
the bottom of the chest. 4704 sq. in. less 18 sq. 
in. = 4686 sq. in. to be covered. The exposed 
face of each corner piece must be covered. The 



Poore's Foundation Course 65 

end of any of the corner pieces is in the form 
of a right angled triangle whose base and per- 
pendicular are each 3". The hypothenuse of 
the triangle is the width of the exposed surface 
of the corner piece. (See sec. 21.) The sum of 
the squares of the two sides that enclose the 
right angle is (3" x 3") plus (3" x 3") = 18 
sq. in. and the square root of 18 sq. in. is 4.24", 
the hypothenuse, or the width of the exposed 
surface of a corner piece. 4.24" x 36" ~ 152.64 
sq. in., the exposed surface of a corner piece; 
and on four corner pieces there is a surface of 
610.5 sq. in. Adding the exposed surfaces of 
ends, sides, bottom and corner pieces, we have a 
total of 14,944.5 sq. in. of surface to be covered 
with tin. 

c. Note the use of dotted lines in the draw- 
ing to show outlines that cannot be seen from 
the view point of the drawing. Assign new 
dimensions to the bins and make a drawing to 
scale, and work the same problem. 

Sec. 65. No. 85 represents a hoisting bucket 
made of ^" sheet steel, to figure its weight 
when filled level full of sand. Study the draw- 
ing until you know the construction of the 
bucket, then make the drawing, using dotted 
lines where proper to use them. 

a. First we will figure the material in the 
bucket. The four sides are made of one sheet 
riveted together with a lap of ij^". The length 
of this sheet is 4 x (24" plus >4") pl^s i>4" 
= 98^ long by 21" wide, containing 2068.5 
sq. in. The bottom is made of a square sheet 
with its four sides bent up so as to form a pan, 



66 Poore's Foundation Course 

a piece having been cut from each corner. The 
bottom of the pan is 24" x 24" (outside) and 
contains 576 sq. in. The sides of the pan form 
a strip 95" long by 2j4" wide (its height in- 
side), containing 213.75 sq. in., which added to 
the bottom = 789.75 sq. in. of material in the 
pan. The top band is 99" long by 2" wide, and 
contains 198 sq. in. Adding the sides, bottom 
and top band together we have 3,056.25 sq. in. 
of sheet steel >4" thick, or 764.06 cu. in. of 
steel. Now we must add the handles or bails. 
We find that the handles are made of a i" 
round bar flattened for 6" on each end, the 
round part between being bent so that the cen- 
ter line of the shaft forms a half circle with a 
radius of 2". This center line is the length of 
the round portion. The diam. of the circle of 
which it forms an arc of 180 degs. (Sec. 19) 
is 4", and its circumference is 4'' x 3.1416 = 
12.566" (Sec. 23,a). Half the circumference, 
or an arc of 180 degs. is 6.283", the length of 
the round portion along its center line. Add to 
this the 6" on each end that has been flattened 
and we have a round bar 18.283" long, and the 
two handles would form a bar 36.566" long by 
i" diam. (Sec. 2y.) (i" x i" x .7854) x 
36.566" = 28.72 cu. in. Adding 764.06 cu. 
in. we have 792.8 cu. in of steel in the bucket, 
weighing 792.8 x .283 lb. = 224.36 lbs., weight 
of bucket. 

&. The inside dimensions of the bucket are 
24" X 24" X 21" zz: 12,096 cu. in., less the sides 
of the pan in the bottom, which we have found 
(a) contains 213.75 sq. in.; and this multiplied 



Poore's Foundation Course 67 

by %'' (thickness) = 53.44 cu. in. 12,096 cu. 
in. — 53.4 cu. in. zn 12,042.56 cu. in. of sand 
contained in the bucket, = 6.969 cu. ft., weigh- 
ing 95 lbs. per cu. ft. = 662.05 lbs. of sand in 
the bucket. Add the weight of the bucket, 
224.36 lbs., and we have a total of 886.4 lbs. in 
the loaded bucket. (See sec. 24,0.) 

c. How can you tell from the drawing that 
the turned-up sides on the bottom are fitted in- 
side and not outside the walls of the sides of 
the bucket? In what way does the drawing of 
the band around the bottom differ from that 
around the top? What does the dotted line at 
the bottom indicate? Change the dimensions 
to 26'' square inside by 23" deep outside; make 
the top band and the bottom turn-up each J^" 
wider, and the radius of the center line of the 
handle 2j^". Draw to scale and figure the 
weight of bucket and load of sand. 

b. Section A A shows a section of the joint 
taken at A A, It shows the rivet in place, and 
is introduced here to show how the joint is 
made and how the rivet is applied. 

Sec. 66. No. 87 shows an ordinary kitchen 
tank, to figure the gallons and weight of water 
it holds. Make the drawing. Note the dotted 
lines and what they mean. The tank is round, 
of uniform diam., and its ends or bases are 
parallel, the lower being concave and the upper 
equally convex. It is a cylinder, 24'' diam. by 
61" long. Its contents are 24" x 24" x .7854 x 
61" =: 27,595.8 cu. in. of water (Sec. 27.) 
weighing .036 lb. per cu. in. (Sec. 24,0) = 
993.45 lbs. of water. Also, dividing 27,595.8 



68 Poore's Foundation Course 

cu. in. by 231 cu. in., we have 115 gals., the 
contents of the tank^ there being 231 cu. in. in 
a gallon. (Sec. 24,a.) 

a. Observe carefully the use of dotted Hues 
and why they are used. Note that the plan is 
broken outside the center line, which may be 
done. Take the dimensions of your own hot 
water tank, make a drawing to scale and find 
the weight of a tank full of water. 

Sec. 67. In No. 84 we have represented a 
stone culvert, to find the number of cubic yards 
of masonry in it. The area of the end of the 
walls of the culvert multiplied by its length 
must give us the contents of the walls. The 
walls are 14" thick (minimum), and any pro- 
jections beyond that thickness on the outside 
are not considered. The arch begins 4'- o" above 
the base, so that the length of the faces of the 
side walls from the base to the arch is 8'-o". 
The length of the face of the base is 4^-0" plus 
14" plus 14" = 6>^'. Adding the length of 
the faces of the side walls we have a face 14^' 
X I %' (or 14") > containing 16.722 sq ft. The 
face of the arch is half of a circular ring whose 
inside radius is 2^-0" and whose outside radius 
is 24" plus 14" = 3 % ". Each of these radii 
describes an arc of 180 degs., or two half cir- 
cles whose diameters are respectively 4'- o" and 
6^' and half the difference of the areas of 
these circles is the area of half the circular ring, 
hence the area of the face of the arch.( See sec. 
47-a.) The difference of the squares of these 
diameters is ^^% sq. ft. less 16 sq. ft. = ^'^% 
sq. ft., which multiplied by .7854 = 18.9368 sq. 



Poore's Foundation Course 69 

ft., or the area of the circular ring; half this 
area is 9.468 sq. ft., the area of half the cir- 
cular ring or the area of the face of the arch. 
Adding the face of the walls below the arch we 
have 26.19 sq. ft., the area of the face of the 
walls of the end of the culvert; this area multi- 
plied by the length of the culvert gives us the 
contents of the walls of the culvert. 26.19 sq. 
ft. X 126 ft. =: 3299.94 cu. ft, or 122.22 cu. 
yds. — ^practically 122% cu. yds. of masonry. 

a. The drawing shows end and side eleva- 
tions, the latter being broken so as to show 
only the ends of the side. Make the drawing 
carefully, giving much attention to details. 
Change the dimensions and make another draw- 
ing to any convenient scale, and find contents 
of the masonry. 

Sec. 68. No. 83 shows a pile of wood drawn 
in perspective (Sec. 11.) The drawing shows 
a lot of wood cut in 30" lengths and piled care- 
fully in three parallel rows abutting each other, 
and supported on one end by the side of a 
building and on the other by a fence. How 
many cords of wood in the pile? Make the 
drawing with care in perspective. The width 
of the pile is 3 times 30" — 7>4'. 

The contents of the pile equals 5'-o" x 7J4' x 
14'- o" :=: 525 cu. ft. But some wood has been 
taken from the pile, and we estimate the de- 
pression to be an average of 12'' deep for a dis- 
tance of 5'-o" along the pile; so it would con- 
tain I'-o" X 2'- 6" X 5'-o'' = 12.5 cu. ft. 525 
cu. ft. — 12.5 cu. ft. = 512 cu. ft., practically, or 
four cords (Sec. 24,a.) 



70 Poore's Foundation Course 

a. Practice sketching and perspective draw- 
ing of objects at hand. Draw a woodpile of 
your own conception and find how much wood 
in it. 

Sec. 69. No. 88 represents a discharge pipe 
of a mine pump, showing an elevation of a side 
of the pipe between the connection for the dis- 
charge elbow and that for the pump elbow, these 
two elbows being shown in broken line. We 
wish to figure the weight of water and the 
pressure to the square inch, either downward 
or in a lateral direction, at the level of the pump 
elbow connection when the water stands in the 
pipe at the level of the discharge elbow con- 
nection. The pipe is 18" in diam. and 280'- o" 
high between these two levels. The area of a 
section of the column of water in the pipe is 
18" X 18" X .7854 = 254.47 sq. in., which mul- 
tiplied by the height in inches (280'- o", or 
3360") = 855,019 cu. in of water, weighing 
.036 lb. per cu. in., or 30,780 lbs., or 15.39 T. 
(Sec. 24-a). The pressure due to gravity^ either 
in a downward direction or in a lateral direc- 
tion, on a square inch of surface at any point is 
the weight of a column of water j" square by 
the height of water above that point. The col- 
umn of water in the pipe being 3,360 in. high, 
the pressure per sq. in. at the level of the pump 
elbow connection is the weight of a column of 
water i" square and 3,360" high, or 3,360 cu. 
in. of water weighing .036 lb. per cu. in. 3,360 
X .036 lb. = 121 lbs. pressure per square inch 
at the bottom or on a square inch of the side of 
the pipe at the level of the bottom. Also ob- 



Poobe's Foundation Coxjbse 71 

serve that the height of water in feet multiplied 
by .^5J = the number of pounds pressure per 
square inch, 280 ft. x 433 1= 121 lbs. pressure 
per sq. in. Note that the number of pounds 
pressure per square inch at any point is 
just a Httle less than half the number of feet 
of water above that point. In any given tank 
the total pressure on the bottom depends upon 
the area of the bottom, and the height of water, 
and is independent of the shape of the tank or 
the quantity of water. The pressure on the 
bottom may thus be greater than the weight of 
water in the tank, or it may be less. Draw three 
vessels as follows: let A be a vessel having 
straight sides, in the form of a cylinder resting 
on one end; let J5 be a vessel in the form of a 
frustum of a cone resting on its larger end (Sec. 
30) ; and let C be a vessel in the form of a frus- 
tum of a cone resting on its smaller end. Let 
all be filled with water to the same height. The 
pressure per sq. in. on the bottom of all three 
vessels will be the same because it depends upon 
the height of water in the vessel ; and this press- 
ure is the same on any square inch and on every 
square inch of the bottom of each vessel. The 
pressure on each bottom, therefore is the press- 
ure per sq. in. multiplied by the number of 
square inches in the bottom. In A this total 
pressure will be just the weight of the water in 
the vessel ; in 5 it will be more than the weight 
of the water; and in C it will be less than the 
weight of the water. You will observe that 
pressure and weight are not necessarily the 
same, although they may be. Think this sub- 



72 Poore's Foundation Course 

ject over carefully, and apply these principles 
to vessels of different shapes. 

a. The walls of the pipe are i^'' thick, and 
the sections 14'- o'^ long. On each end of each 
section there is a flange for connecting the sec- 
tions; and these flanges are i^" thick and have 
an outside diam. of 26''. Figure the weight of 
the pipe. The inside diam. of the pipe is 18"; 
the outside diam. is 21". The contents of the 
walls of the pipe = the area of the circular ring 
shown by a section of the pipe, multiplied by 
the length of the pipe. The square of 2i"mmus 
the square of 18" x .7854 — 91.8918 sq. in., the 
area of the end of the pipe, not including the 
flange. 91.8918 sq. in. x 3360" (length) =^ 
308,756.45 cu. in. of cast iron in the pipe, not 
including the flanges. Regarding the flanges as 
separate from the pipe, each flange has an out- 
side diam. of 26" and an inside diam. of 21". 
The area of the face of one side = the square 
of 26" minus the square of 21" x .7854, or 
184.57 sq. in. 184.57 sq. in. x i%" (thickness 
of flange) = 323 cu. in. of metal in one flange. 
There are twenty sections of pipe, each having 
two flanges, or forty flanges. 323 cu. in. x 40 
=: 12,920 cu. in. of cast iron in the flanges. 
Adding 308,756.45 cu. in., we have 321,676.45 
cu. in. of cast iron, weighing (Sec. 24-a) 
83,635.9 lbs., or 41.8 tons. 

b. Beginning at the first joint below the top, 
what is the water pressure per square inch at 
each joint, the pipe being full? 

Sec. 70. No. 86 shows a plan, an end eleva- 
tion and a side elevation of a brick powder 



Poore's Foundation Coubse 73 

house with stone foundation walls, to figure 
cubic yards of masonry in foundation, and num- 
ber of brick in the walls of the building. Make 
drawing to scale. Note dotted lines and why 
used. In drawing pay careful attention to de- 
tails. 

a. The base of the foundation extends 6" 
on either side of the i8'' wall above it. (See 
front elevation). This base is 2'- 6" wide and 
6" high. Figuring the length of this base across 
the front and the rear, and on the sides between 
the end bases, we have 13'- o" plus 13'- o" plus 
6'- o'^ plus 6'- o" = 38'- o" long by 2j4' wide by 
>^' high = 47.5 cu. ft. in the base. Now, 
figuring the wall above the base in the 
same way, we have 12'- o" plus 12'- o" 
plus 7'-o" plus 7'-o" =1 38^-0" long by 
4'-o" high by i>4' wide =: 228 cu. ft., which 
added to 47.5 cu. ft. = 275.5 c^- f^- =^ i^-^ c^- 
yds. of masonry in the foundation walls. Where 
walls are exposed and corners have to be dress- 
ed, stone masons figure the contents of the wall 
by taking the outside measurements from cor- 
ner to corner as the length of the wall on any 
side or end. 

b. The brick wall is 12^" thick and has 21 
bricks to the square foot (See sec. 24-6. ). The 
distance around the wall of the building less 
four times the thickness of the wall gives us 
the actual length of wall; and the length multi- 
plied by the height gives us the surface for this 
length. The distance around is 44'- o". 4 x 
12^- = 51- = 4%\ 44^-0- - 4%^ = 39%' 
~ 39.75' .Multiplied by the height (io'-o"),we 
have 397.5 sq. ft. to which we must add the ga- 



74 Poore's Foundation Course 

bles. They are two triangles having a 12'- o" 
base and altitude of 3'-o". The area is half 
their product, or 18 sq. ft., and in two gables 
there are 36 sq. ft. Adding to 397.5 sq. ft., we 
have 433.5 sq. ft. Now deduct the door. It is 
3'- 6" wide and 5^-3" high to the foot of the 
arch. We will add half the height of the arch 
for an average height, and we have 5'- 6" for 
the height of the door. Its area will be 3.5' x 
5.5' z=z 19.25 sq. ft. Subtracting from 433.5 sq. 
ft. we have a wall surface of 414.25 sq. ft. hav- 
ing 21 bricks to a square foot, or 8,700 bricks. 

c. In figuring the cost of laying bricks at so 
much per thousand, ordinary doors and win- 
dows are not deducted ; neither is the thickness 
of the wall taken off at the corners ; the distance 
around the outside of the wall is considered the 
length of wall. Make a drawing of some fa- 
miliar building and figure the number of bricks 
in the walls, and also how many thousand bricks 
would be charged for in the bricklayer's bill for 
laying the walls. 

Sec. 71. In No. 100 we have a comparative 
illustration of roof angles, showing those most 
commonly used. At the top is shown a 45 deg. 
pitch, in which the rise is 12" in 12". The 
angle of 45 deg. is the angle formed by the roof 
line and the horizontal across the building — ^the 
angle at the eaves. The roof pitch is designated 
either by the measure of this angle or by the 
rise per foot; that is, the number of inches of 
rise for every foot of length of the horizontal 
from the outside edge of the building to the 
center line. In the case of a 45 deg. pitch, if 



Poore's Foundation Course 75 

the length of the horizontal from the outside 
edge to the center line is 12'- o", the perpendic- 
ular from the center line to the apex will be 
12'- o". In the case of the 8" pitch the perpen- 
dicular will be 8'- o". In the 30 deg. pitch the 
angle with the base or horizontal is an angle of 
30 deg. In a 6" pitch the rise is 6" for every 
foot of length of the horizontal to the center 
line; that is, the perpendicular upon the center 
line will be half the length of the horizontal to 
the center line. In a 22>4 deg. pitch the angle 
with the horizontal is one-fourth of a quadrant, 
or 22j^ deg. A Ys pitch means that the perpen- 
dicular is one-third the length of the horizontal 
to the center line ; and a }i pitch means that the 
perpendicular is ^ the horizontal. 

a. The length of the horizontal to the center 
line being known, and the pitch being decided, 
you determine the length of the perpendicular 
from the center line to the apex of the roof ; and 
with the perpendicular and the horizontal, or 
base, you can find the hypothenuse (Sec. 21), 
and that will be the length of the rafter from 
the apex to the outside end of the horizontal; 
add the length of the projection for the eaves 
and you will have the length of the rafter. In 
small buildings, carpenters generally find the 
length of the rafter by making a pattern or tri- 
angular frame, the base and perpendicular hav- 
ing been determined as above; they can then 
measure the hypothenuse and get the length of 
rafter. 

Sec. y2. Nos. loi-a to 108 show a general 
drawing of a 6" pitch wooden roof truss. It is 



76 Poore's Foundation Course 

a general drawing because it shows the parts in 
place. No. loi is the chord of the truss; No. 
102 is a rafter, of which there are two ; No. 103 
is a strut, of which also there are two; No. 104 
is the king bolt, and 105 is a square 
nut; No. 106 is a strut bolt; No. 107 
is a toe bolt, there being two of each. No. 108 
is a square nut. Nos. 104 and 105, and Nos. 
106, 107 and 108 are shown also in detail. Nos. 
3-75, 3-76 and 3-77 are cast iron washers, and 
these numbers refer to Nos. 75, 76 and yy on 
Sheet No. 3, where the items are shown in de- 
tail. Likewise No. 3-78 and 3-79 refer to Nos. 
78 and 79 on Sheet No. 3. No. 78 is a peak 
washer and No. 79 is a toe plate. Note care- 
fully the details of these numbers on Sheet No. 
3; draw washers and toe plates of different di- 
mensions to scale. Draw the completed truss — 
a general drawing, all parts being shown in place. 

Sec. 73. No. 112 shows a steel taper pin, side 
and end views. It is introduced here principally 
for the purpose of affording a practical illustra- 
tion of the use of converging shading lines rep- 
resenting a convex surface. They should be 
drawn with care. Note the difference between 
the shading lines on No, 112 and those on No. 
69. No. 112 represents a frustum of a cone 
(Sec. 30) and the shading lines are not parallel. 
Draw frustums of different cones, and carefully 
draw the shading lines to show where the shad- 
ow lies on the frustum, and that the surface is 
convex. 

Sec. 74. We come now to the screw, always 
an interesting subject in drawing. A cylinder 



Poore's Foundation Course 77 

grooved or threaded in an advancing spiral on 
its outer surface, also a hollow cylinder having 
such a groove or thread on its inner surface are 
forms of the screw. 

The former is ordinarily called a screw and 
the latter a nut. The threads of the nut are 
adapted to turn on the threads of the screw. 
Nos. 104, 105, 106, 107 and 108 present common 
forms of screws and nuts. Nos. 114 to 122 in- 
clusive show various styles of machine screw 
heads and points with their respective names. 
You should become familiar with them all. Any 
desired point may go with any desired head, so 
that from the styles shown a great number of 
combinations of heads and points may be made. 

a. Nuts are ordinarily either square or hexa- 
gon shaped. The end view of No. 124 shows a 
^" hexagon nut, and also in broken line a 
^" square nut for the purpose of comparison. 
Nuts are designated by their greatest inside 
diam. ; that is, by the outside diam. of the screw 
which they were made to fit. Notice that the 
distance over flats is the same on both nuts. It 
is a standard dimension and a wrench that will 
fit one nut will fit the other. The square nut 
requires a circular turning space 1.502" in diam., 
while the hexagon nut will turn in a circular 
space 1.227" ^^ diameter. In No. 138 at the top 
of Sheet No. 5 there are shown two views of a 
special form of hexagon nut called a safety nut. 
Observe that a hollow cylinder J^" long is turn- 
ed on one end of the nut, and that three slots ^" 
wide and i/^" deep are made diametrically across 
the end of the hollow cylinder. In the end of 



78 Pooee's Foundation Course 

the screw or bolt there is made a round hole to 
receive a cotter pin or split pin (No. 139). When 
the safety nut is in place on the bolt the cotter 
pin is inserted in one of the slots in ihe 
nut and is pushed through the hole in the bolt, 
thus preventing the turning of the nut. In No. 
128 you see a pair of hexagon lock nuts. They 
may be locked in any position on the screw by 
holding one nut and turning the other tightly 
against it. Another name for them is jamb nuts. 

b. There are three styles of thread ordinarily 
used in making screws. They are the U. S. 
standard thread, as shown on the left end of 
No. 127; the square thread, as shown on the 
right end of 127; and the 29 deg. Acme thread, 
as shown on No. 140 and No. 141. The U. S. 
standard thread is the thread most commonly 
used. In U. S. standard threads the number of 
threads on each inch of the screw depends upon 
the diameter of the screw, the number increasing 
as the diameter decreases, as follows : i" diam. — 
8 threads per inch ; %" diam. — 9 threads ; ^" di- 
am. — 10 threads ; ^" diam. — 1 1 threads ; J4" di- 
am. — 13 threads ; ^" diam. — 16 threads ;%6" di- 
am. — ^18 threads ; J4"d!iam. — 20tjhreads. Automo- 
bile manufacturers have agreed upon a special 
standard of threads per inch for use in automobile 
construction, requiring finer threads and more 
threads per inch. No. 123 represents a machine 
stud, threaded at both ends. It is ^" in diam. 
and there should be 11 threads per inch, U. S. 
standard. No. 124 is the nut in place. The 
thread on the other end is cut just slightly larger 
than U. S. standard, so that it may be screwed 



Poore's Foundation Course 79 

very solidly into a U. S. standard tapped hole. 
No. 125 is a machine bolt differing from a 
machine stud in that it has a head on one end. 
Notice that the threads on Nos. 123 and 125 are 
not pictured; they are simply indicated. This 
method of representing threads is very generally 
used in order to save time. The number of lines 
and spaces used to represent a thread bears no 
relation to the number of threads per inch. No. 
136 is a special automobile bolt^ 24 threads pef 
inch. The drawing simply shows that the bolt 
is threaded ; the lines do not indicate the number 
of threads. Observe that the head on No. 125 is 
square. The angle or corner of the nut is rep- 
resented as being toward you. The square nuts 
on Nos. 104 and 106 are drawn the same way. 
Please note that carefully. Now observe No. 
126. It is a representation of a square nut 
shown as if you were looking directly at one of 
its sides or faces. Square nuts or heads are of- 
ten drawn in this manner. 

c. No. 127 shows a cross feed screw of a 
lathe, having a square thread on the right hand 
end. Each thread is )^" wide, and each groove 
is >^" deep and J^" wide; a section of either 
thread or groove would be >^" square, practical- 
ly, and a thread and a groove occupy J4" of the 
length of the screw; so we say there are four 
threads to the inch, measuring the width of a 
thread and a groove for the width of a thread. 
At the end of the square thread there is a shoul- 
der. That this shoulder is round is shown by the 
fact that its one dimension is given as a "diam- 
eter"—"!^'' dia." Note that carefully. That 



80 Poore's Foundation Course 

the shank next to the shoulder is round is shown 
also by the dimension ''J^'' dia." A U. S. stan- 
dard thread is cut on a portion of this %" 
shank. There are nine threads to the inch, the 
portion of thread shown having 4^ threads to 
a half inch. The standard for a ^" screw is 
nine threads per inch. 

(&) No. 128 shows two hexagon nuts 
working one against the other and called 
jamb nuts or lock nuts. By holding the 
nut nearest the shoulder and tightening the 
other against it, the distance to the shoulder may 
be definitely fixed and regulated so that the part 
of the mechanism that is to be held between the 
nuts and the shoulder may be held without any 
lost motion and yet not so tightly that the screw 
may not turn readily. Beyond the U. S. stan- 
dard thread the shaft is turned down to a diam. 
of ^", and four flat faces are cut on it, making 
it almost square, to receive a handle or a hand- 
wheel, as is shown by the section. 

d. No. 140 shows a 29 deg. Acme thread cut 
on a ij4" shaft, there being two threads to the 
inch, and the thread is called a right hand 
thread. No. 141 shows a left hand thread of 
the same kind. All screws are made with right 
hand threads unless left hand threads are speci- 
fied for some particular use. Note that No. 141 
shows the screw without the shading lines, which 
is often done to save time. 

e. No. 66, Section AA, on Sheet No. 3, 
shows a section of a pipe thread. Iron pipes are 
regularly threaded so as to form a tapering 
screw ; that is, the diameter of the screw becomes 



Poobe's Foundation Course 81 

gradually less as the end of the pipe is ap- 
proached. This taper is for the purpose of mak- 
ing tight joints. No. 66 shows it slightly exag- 
gerated. 

/. No. Ill shows a sectional view of a U. S. 
standard thread. The distance between centers 
of adjacent threads is called pitch, and in the 
view shown the pitch is ^", or two threads to 
the inch. The sides of a U. S. standard thread 
form an angle of 60 deg. with each other. The 
thread is flat, top and bottom, and the width of 
this flat is % of the pitch. 

g. The thread diagram, No. 113, is full size, 
and shows how to draw a U. S. standard thread. 
The screw is to be ij4" in diam. ; there must 
therefore be seven threads to the inch. The 
pitch must be Vr". To draw the thread, first 
draw the center line c; then with your bow- 
dividers open ^" locate two points that distance 
on either side of the line c, and through these 
points draw the lines a and b parallel to the line 
c. They will be ij4" apart; and this dimension 
represents the outside diameter of the screw. 
Now take your bow-dividers and adjust them 
so as to divide an inch into seven equal parts. 
The length of one of these parts is the pitch of 
the thread, W\ With the bow-dividers thus 
adjusted divide the line a into equal spaces. In 
No. 113 we have some of these division points 
shown greatly exaggerated. In practice the 
puncture made in the paper by the bow-dividers 
should be a sufficient mark to indicate the divi- 
sion points. Now see No. 20, Sheet No. i. Se- 
lect your triangle having in it an angle of 30 



82 Poore's Foundation Course 

deg. Set its shorter base on the upper edge of 
the T-square, and draw the short lines e 
through all the division points in the line a. 
Then turn over the triangle and drav^ the short 
lines d through the same division points. The 
intersections o of these short lines e and d in 
the division points in the line a mark the centers 
of the tops of the threads. Their intersections 
/ below the line a mark the centers of the root 
of the thread. Now from one of these root 
centers / draw a line v perpendicular to the line 
h; it will intersect it at g. From gr as a starting 
point divide the line h into equal spaces repre- 
senting the pitch of the thread, the same as you 
divided the line a. Through these division points 
(igr) in line h draw the short lines h and k in 
like manner as you drew e and d. Then draw the 
lines p from the intersections o to the intersec- 
tions g; and also draw the lines m from the in- 
tersections / to the intersections r. Having drawn 
all of these lines as directed in pencil you are 
ready to ''ink in." Ink the lines e and d only 
from intersections o to intersections f, and ink 
the lines h and k only from intersections g to 
intersections r. Ink the lines p and the 
lines m, and also ink the center line c\ You 
have now a thread drawn to scale, full size, and 
you are ready to erase the pencil lines. If you 
inked only the lines, a, h, c, />. and m, you would 
have an indicated thread drawn to scale, full 
size. 

h. In drawing an indicated thread you will 
find it very helpful to draw the pencil lines s 
and t at such distances from a and b respectively 



Poobe's Foundation Course 83 

as would represent the depth of the thread. The 
lines a and b become the Hmits of the lines p 
representing the top of the thread, and the lines 
s and t become the limits of the lines m repre- 
senting the root of the thread. The lines s and 
t should not be inked; they should be erased af- 
ter inking the lines a, b, c, p and m. As the ob- 
ject in indicating a thread is usually to save 
time, it may or may not be drawn to show the 
true number of threads to the inch. But in any 
event you must first determine the number of 
threads you will show to the inch to make your 
work look well, and divide the line a into equal 
spaces accordingly. You will then proceed as 
above to locate the lines p and m, having first 
drawn the lines .y and t through the intersections 
/ and r respectively. 

Sec. 75. Nos. 143, 144 and 145 present an- 
other use of the spiral in the form of spiral 
springs. No. 143 is a conical compression 
spring, made to occupy the least possible space 
when compressed. It may be compressed until 
its height or length is no more than the thickness 
of the wire of which it is made. 

a. No. 144 is the most common form of 
spiral compression spring, called a cylindrical 
spiral. It is an open spiral. It may be com- 
pressed until the coils lie one upon another. The 
virtue of Nos. 143 and 144 lies in their ability 
to resist compression, and to return to their 
original position as the compressing force 
is released. The property which tends 
to cause a return to the original position is 
called elasticity. 



84 Pooee's Foundation Course 

b. No. 145 presents a closed cylindrical spiral. 
It tends to resist being pulled open, and to re- 
turn to its closed position upon being released. 
It is called a tension spring. A common use of 
a spiral tension spring is in an ordinary spring 
balance. Its virtue lies in its elasticity, or its 
ability to return to a closed position upon being 
pulled open. 

Sec. 76. By way of illustrating a subject re- 
quiring greater detail we have introduced draw- 
ings of a familiar bit of modern mechanism — an 
automobile piston, connecting rod and crank 
shaft. No. 129 shows the piston, presenting a 
side view showing the end of one of the wrist 
pin bearings. The piston is round, turned all 
over. There are three grooves each J4" wide, 
turned around the outside of the piston to re- 
ceive the expansion rings shown in No. 130. We 
have two views of the ring. Observe that the 
inside circumference and the outside circum- 
ference are not concentric; their centers are not 
at the same point; they are eccentric, and the 
ring is thicker at one side than at the other. No- 
tice that there is a cut through the ring where it 
is thinnest, and that this cut is %6 " long and 
that it passes through the ring from one side to 
the other at an angle of 45 deg. with the side 
line of the ring. The grooves around the piston 
are machined to gauge — to an exact width; and 
the rings are machined to an exact width to cor- 
respond. The grooves are slightly deeper than 
the thickness of the rings at their thickest part. 
The rings are sprung open and slipped over the 
end of the piston and moved along until each 



Poobe's Foundation Course 85 

springs into its groove. Then they are com- 
pressed as the piston enters the cylinder; so 
that their tendency to expand keeps the outer 
surface of the rings in close contact with the 
inner surface of the cylinder. No. 131 presents 
an end view of the piston showing in outline the 
brass bushings that receive the wrist pin (shown 
in No. 132 at the upper left hand corner of the 
sheet). The wrist pin is finished all over and 
is fitted carefully into the bushed bearings made 
to receive it in the piston. Note that there 
is a concave cut made in one side of it, and that 
it has a screw slot in one end. The latter is for 
the purpose of being able to turn the wrist pin 
in its bearings so as to bring the concave cut to 
the right place so that it will be occupied by the 
clamp bolt when in place, thus preventing the 
wrist pin from moving toward either end, and 
keeping it from contact with the inner wall of 
the cylinder. 

a. Nos. 133, 134, 135, 136, 137 and 138 show 
a general drawing of the connecting rod and 
also detailed drawings of its parts. We have a 
side elevation of the connecting rod with a sec- 
tion, a plan, an end elevation and another sec- 
tion. No. 133 shows the plan of the connecting 
rod, and below it is a side elevation. They show 
on the wrist pin end of the connecting rod the 
chamber bored out for the wrist pin and the 
cylindrical boss bored and tapped for the clamp 
bolt (No. 136), and also a Vie" milled cut 
through the upper wall of the wrist pin cham- 
ber and through the boss. This milled cut al- 
lows the clamp bolt, when in place, to be drawn 



86 Poore's Foundation Course 

up so as to pull together the walls of the cut, 
tightening the chanber around the wrist pin like 
a clamp. The side elevation shows also how 
the clamp bolt occupies the concave cut in the 
side of the wrist pin and prevents its moving in 
the direction of either end. In the other end 
of the connecting rod is the crank shaft bearing 
showing the babbitt in the bearing and the cap 
bolts (No. 137) and the safety nuts (No. 138) 
in place. The side elevation, plan, section and 
end view show four different views of this end 
of the connecting rod. The bearing is in two 
parts, the division being along the line of the 
vertical center line as shown in the drawing. 
The detached piece is called the cap, and when 
the journal of the crank shaft is in place in its 
connecting rod bearing the cap is securely fas- 
tened on by means of the two cap bolts No. 137, 
shown also in detail at the top of the sheet. One 
side of the heads of the cap bolts is cut off so 
that the straight edge thus formed prevents the 
cap bolt from turning by its contact with the 
shoulder shown in the section of the connecting 
rod. The safety nuts (See Sec. 74-a) are 
shown in place and with the cotter pins (No. 
139) also in place, so that the nuts cannot turn 
and become loosened. The section D-D is a sec- 
tion of the cap, as though the lower half of the 
cap were cut off at D-D and you were looking 
up at the exposed surface remaining. No. 134 
is the cap and No. 135 is the babbitt within. The 
curved dotted line shows a shoulder made by 
providing clearance for the nuts No. 138. 
b. No. 142 shows an automobile crank shaft 



Poore's Foundation Course 87 

to go with the connecting rod shown. It has 
three main journals, the two on the ends being 
each i^" diam. x 3" long, while that in the 
middle is ij^" diam. x 2j4" long. There are 
four connecting rod journals each 1%'^ diam. 
X ij^" long, having a two inch throw, which 
is just half the piston stroke. The center lines 
of all the journals are within the same plane. 
This is shown by the end views. The view from 
the left hand end shows the lines of the long 
arms and of the short arms, while the section at 
A'A shows the sectional outline of the long 
arms and that at B-B shows the sectional outline 
of the short arms. Both end views show the 
coupling flange on the right hand end. It is 
made to be attached by six bolts to a gear coup- 
ling. The dotted circles indicate the main jour- 
nals and their shoulders. The two small dotted 
circles in the left hand end view show the out- 
lines of the connecting rod journals within the 
lines of the shoulder or flange at each end of 
the journal. 

Sheet No. 5 is full of practical work in draw- 
ing and in reading drawings. You should not 
leave it until you can draw well any number on 
it. The spirals — the screw thread and the spiral 
springs should receive plenty of attention. When 
you have mastered the work on this sheet you 
will be able to draw any screw thread, double 
threads excepted. Study the piston, connecting 
rod and crank shaft drawings until you abso- 
lutely understand every line and why it is there 
— until you can see the mechanism before your 
eyes by looking at the drawings. Then you will 



88 Poore's Foundation Course 

have read the drawings. It will be fine practice 
to present this subject in full size and half size 
drawings. Draw from original subjects. Se- 
lect simple subjects and make different views 
of them. Gradually take up subjects having 
more detail. Do your best on everything you at- 
tempt to draw and do not leave it until you know 
that you have done it well. In the meantime do 
not fail to apply your mensuration wherever you 
can get data on which to base a problem. The 
mathematical end of your work is very important. 
Drawing without mathematics makes a one 
handed man — not very useful anywhere. Com- 
bine the two with ability to think and unceasing 
application, and you will overcome all obstacles. 
Sec. 77 . We come now to consider the lever, 
because it enters into so many mechanical con- 
structions. There are three classes of levers, 
and three elements to be considered in connection 
with the lever: — the fulcrum is the point about 
which the lever turns, as F (No. 164) ; the 
weight or resistance to be moved or overcome, 
as W; and the power or force applied, as P. 
When the fulcrum is between P and W the lever 
is of the first class. Resistance or weight 
and power or force are measured in pounds; 
and when the power multiplied by its distance 
from F equals the weight multiplied by its dis- 
tance from Fj the power and the weight will 
balance each other. Therefore P x distance P-F 
=r W X distance W-F; and three of these four 
elements being known, the fourth can be found. 
In No. 164, if the weight of the boy is 100 lbs., 
and the distance from the middle point between 



Poore's Foundation Course 89 

his hands to the point where the crow bar rests 
on the block is 36", and the distance between 
this point (fulcrum) and the point where the 
crow bar touches the big stone is 6", 100 lbs. 
X 36 =: 3,600 lbs. ; divide by 6 and we find that 
the boy can exert a lifting force of 600 lbs. To 
move the stone one inch the boy's hands will 
have to travel six inches, and the power will 
move six times as fast as the weight ; hence also, 
the power multiplied by the distance it travels 
must equal the weight multiplied by the distance 
it moves, and three of these elements being 
known, the fourth can be found as in the above 
problem. 

a. No. 165 represents a lever of the second 
class, the weight being between the power and 
the fulcrum. Suppose the boy exerts a lifting 
force of 100 lbs., and the distance from the mid- 
dle point between his hands (P) to the point 
where the crow bar rests on the ground (ful- 
crum) is 42", and the distance from that same 
point (fulcrum) to the point where the stone 
rests on the crow bar is 6"; 100 lbs. x 42 = 
4,200 lbs., and dividing by 6 we find that the 
boy can exert a lifting force of 700 lbs. on the 
stone. To move the stone one inch the boy's 
hands will have to travel seven inches, and the 
power will move seven times as fast as the 
weight. 

b. When the power is exerted between the 
weight and fulcrum the lever is of the third 
class (No. 166). Suppose the boy wishes to lift 
a two-pound fish from the water; the weight is 
where the line is attached to the rod ; if he holds 



90 Poobe's Foundation Coukse 

his right hand stationary it becomes the ful- 
crum; and if he lifts with his left hand it be- 
comes the power. If the distance from fulcrum 
to weight is 42", and from fulcrum to power is 
6", then the weight, 2 lbs., x 42 = 84 lbs., and 
dividing by 6 we find that the left hand must 
exert a lifting force of 14 lbs. to lift that 2- 
pound fish. In raising the left hand (power) 
one inch the fish (weight) will be moved seven 
inches. 

c. In levers of the first and second classes 
power is gained and motion is lost; in levers of 
the third class power is lost and motion is 
gained. Look around you for examples of the 
different kinds of levers. They are everywhere. 
Consider their application. Make up problems 
and work them out. A pair of scissors is an ap- 
plication of levers of the first class. A lemon 
squeezer is an application of levers of the second 
class. Revolve a lever about its fulcrum; the 
power would describe a circle, and the weight 
would describe another circle; and so we have 
the principles of the lever applied in wheels, 
pulleys, gears, etc. 

Sec. 78. Pulleys are wheels used to transmit 
motion and power by means of belts. No. 153 
represents the smaller pulley as being a driving 
pulley communicating motion and power to the 
larger. Motion or speed will be lost and power 
will be gained. If the driving pulley is 36" in 
diam. and runs 100 revolutions per minute, at 
what rate of speed will it drive a 48" pulley? 
100 X 36, and this product divided by 48 gives 
us the number of revolutions per minute of the 



Poore's Foundation Course 91 

48" pulley, or 75 revolutions. To find the speed 
of a ''follower," or following pulley, multiply the 
r, p. m. (fet'olutions per minute) of 
the driving pulley by its diameter and di- 
vide by the diameter of the follower, Th e 
result is the r, p, m. of the follower. Likewise 
to find what must be the diameter of a follower 
to give it a certain speed, multiply diam, of driv- 
ing pulley by its r. p, m. and divide by the re- 
quired r. p. m. of the follower. The result will 
be the diam. of the follower. No. 154 represents 
a train of pulleys introduced for the purpose of 
illustrating their use in transmitting increased or 
diminished power and motion, and for changing 
the direction of the latter. Let the 48'' pulley A 
be the driving pulley running 100 r. p. m. It 
drives a 36" pulley B, which we find (See above 
rule.) will run at the rate of 133)^ r. p. m. With 
the pulley B there are the 30" pulley C and the 
8" pulley D, all on the same shaft; and being 
tight pulleys (turning with the shaft), they all 
have the same r. p. m. The pulley D therefore 
makes 133^ r. p. m., and it drives a 36" pulley 
H. 8 X I33>^, divided by 36 = 29.6r. p. m. of 
pulley H. On the same shaft with the pulley B 
an 8" pulley F runs at the same rate and drives 
a 24" pulley G with a reversed motion. Show 
by the same method that the pulley G runs at 
the rate of 9.8 r. p. m., and that the 8" pulley H 
revolves with it in the same direction and at the 
same rate. Now going back to the 30" pulley C, 
it makes 1333^ r. p. m. and drives an 8" pulley 
/. Applying the rule we find that the pulley / 
makes 500 r. p. m. The 24" pulley / with it 



92 Poore's Foundation Course 

drives an 8'' pulley K giving it a speed of 1500 
r, p. m. The idler pulley L does not transmit 
motion or power. It is simply a belt tightener, 
capable of being moved against the belt or away 
from it. 

a. If the driver is an 18" pulley making 300 
r. p. m., what size pulley must be put on the 
line shaft to give it 250 r. p. m. ? (See rule 
above). 18 x 300, divided by 250 = 21.6" 
diam. From this same line shaft running at the 
rate of 250 r. p. m., it is desired to run a jointer 
3,500 r. p. m. The pulley on the jointer is 5" 
in diam. Select a pulley of suitable size to be 
the driving pulley on the line shaft, and then find 
what size follower and driver on the counter- 
shaft will give the required speed approximately. 
You can find plenty of practical examples. Re- 
duce them to figures. Arrange a mill room, 
starting with an engine running 250 r. p. m., and 
put in a line shaft and counter shafts with ma- 
chines. Assign desired speeds to the machines, 
and then arrange your pulleys so as to secure the 
desired speeds, while at the same time trying 
to avoid running your counter shafts at high 
speeds. 

b. In studying levers we learned (Sec. 77'c) 
that when motion is lost power is gained, and 
when motion is gained power is lost. Since pul- 
leys and gears are applications of levers, the 
same principles apply. The power transmitted 
is in an inverse ratio to the motion transmitted. 
In No. 153 the driver makes 100 r. p. m. and the 
follower makes 75 r. p. m. The motion trans- 
mitted is '^%oo of the motion of the driver, or 



Poobe's Foundation Course 93 

^4. The power transmitted to the follower is 
io%5 or i^ times the power of the driver; 
that is, a certain power exerted upon the shaft of 
the driver will exert i^ times as much 
force upon the shaft of the follower, due 
to the advantage of the leverage in the larger 
wheel, but in gaining that advantage the wheel 
will run slower and motion or speed is lost. 

Sec. 79. We come now to consider the trans- 
mission of motion and power by means of gears 
— gear-wheels or cog-wheels (Nos. 155 to 163). 
The usual form of a gear is that of a wheel 
having transverse teeth or cogs on its face, made 
to mesh with similar teeth on another gear. Per- 
fectly formed teeth are made to roll upon each 
other with little or no sliding friction when gears 
are running in mesh. This requires that the face 
of the tooth should have just the proper curve. 
When gears are in mesh the depth to which the 
teeth of one enter between those of the other is 
called working depth, (See No. 155 — Gear Dia- 
gram). G is the working depth of the tooth. 
The bottom, of the trough between the teeth is 
spoken of as the bottom of the tooth. You will 
notice that when the teeth are in mesh the end 
of the tooth of one gear does not quite touch the 
bottom of the tooth of the other gear. The space 
between is called clearance; also, when the 
working surfaces of teeth in mesh are in con- 
tact, between the opposite or idle surfaces there 
is a slight space, and this space is likewise called 
clearance. In No. 155 these two clearances are 
shown at F. and F. Divide the working depth 
of a tooth into two equal parts as measured 



94 Poobe's Foundation Course 

along its center line, and draw a circle from the 
center of the gear through this division point. 
This is the pitch circle of the gear, as H. When 
two gears are in mesh their pitch circles just touch 
but do not intersect each other. Note that the 
pitch circles of the gears shown in mesh in the 
drawing are always tangent to each other — they 
touch but do not intersect. The distance be- 
tween the pitch circle and the end of a tooth, 
measured along a radius of the gear, is called 
addendum; and in practice this name has come 
to be applied to that part of the tooth which lies 
outside of the pitch circle. In like manner that 
part of the tooth lying within the pitch circle is 
called dedendum. In No. 155, A is the adden- 
dum and B is the dedendum. The distance 
measured along the pitch circle from the center 
line of any tooth to the center line of the next 
tooth is called circular pitch. E represents the 
circular pitch of the teeth of the diagram. 
Thickness of tooth is the thickness of any tooth 
measured along the pitch circle, as is shown at 
D. The diameter of the pitch circle is called 
pitch diameter, as I. The outside diameter is 
the diameter of a circle drawn from the center 
of the gear and passing through the end lines 
of the teeth, as J in No. 155; or, it is the diam- 
eter of the gear from end of tooth to end of 
tooth, as J in No. 156-a. The distance measured 
along a radius of a gear from the end of a tooth 
to the bottom of the tooth, or from the circle of 
the outside diameter to the bottom of the tooth, 
is called whole depth of tooth, as is shown at C 
in No. 155. The working surface of the tooth 



Poore's Foundation Course 95 

outside of the pitch circle is called the face of 
the tooth; and that part of the working surface 
of the tooth inside of the pitch circle is called 
the flank of the tooth. All of these various 
terms and the relation of the things they repre- 
sent should be carefully learned. 

a. Nos. 156-a, 157, 158, 159, 156-&, 160 and 
156-^ show a train of gears, of which 160 and 
156-^ are not in the same plane as the others; 
they are behind 159 and 156-6. 156-6 and 156-^ 
are loose gears turning on journals; they do not 
turn shafts. Their office would be to transmit 
motion and power to other gears in the train 
not shown. 159 and 160 are on the same shaft 
and both are tight gears, so that the gears and 
the shaft all turn together. The key-way is 
shown, indicating that they are to be keyed 
tightly on the shaft. 156-a and 158 are also 
tight gears. 157 is a gear cut on a shaft, as is 
indicated by its being shown in section. The sec- 
tion A A is taken ^t A A through the vertical 
center line of 159 and 156-6. The section B, B. 
is taken 3Lt B B through the center lines of 160 
and 156-^. It is an inclined section represented 
in a vertical position in order to show the gears 
in proper relation. The journal for 156-^ is 
shown in place. That it is a journal and not a 
tight shaft is shown by the crossed diagonal 
lines, wihch are always used to indicate journals. 
Likewise the shaft in 159 and 160 is shown to 
be a tight shaft by the absence of the diagonal 
lines, and also by the key-way shown on the side 
of the shaft. 

6. The gears in the drawing are half size. 



96 Poore's Foundation Course 

The diameter of the pitch circle of 158, or its 
pitch diameter, is 3"; the pitch diam. of 159 is 
6". Their gear centers are 4j^" apart, or half 
the sum of their pitch diameters. The gear 
centers of 159 and 156-& are 5" apart; and the 
gear centers of 156-a and 157 are 3" apart. In 
156-a, having a pitch diam. of 4", there are 16 
teeth; in 157, pitch diam. 2", there are 8 teeth; 
in 158, pitch diam, 3", there are 12 teeth; in 
159, pitch diam. 6", there are 24 teeth; in 160, 
pitch diam. 3j^", there are 14 teeth; in 156-b, * 
pitch diam. 4", there are 16 teeth; and in 156-4?, 
pitch diam. 4", there are 16 teeth. It will be 
seen that in each gear in the train the number 
of teeth is four times the pitch diameter, or 4 
teeth for each inch in the pitch diameter. In 
any gear the number of teeth per inch of pitch 
diam. is called diametrical pitch; so in this train 
of gears the diametrical pitch is 4 ; which means 
that the number of teeth is 4 x the pitch diam. 
In any train of gears the diametrical pitch must 
be uniform in all the gears; and what this di- 
ametrical pitch should be in any train of gears 
depends upon the circumstances and conditions 
of their use. 

c. In this train of gears we will assume that 
158 is the driving gear. 158 transmits motion to 
157, and 157 in turn transmits motion to 156-a. 
Observe the direction of the motion of 158 as 
shown by the arrow, and note that when 158 
turns it will turn 157 in the opposite direction; 
157 in turn will impart motion in the opposite 
direction to 156-a^ and the direction of 156-a 
will be the same as that of 158. Thus we see 



Poore's Foundation Course 97 

that numbers one, three, five, etc., in a train of 
gears, or the odd numbers, all turn in the direc- 
tion of No. one — ^the driving gear; while the 
even numbers of gears turn in the opposite di- 
rection. 

d. The pitch diameter of 156-a is the same 
as that of 156-fe, and they each have 16 teeth. 
158 transmits motion to 156-a through 157 hav- 
ing 8 teeth; 158 also transmits motion to 156-&' 
through 159 having 24 teeth. 158 has 12 teeth. 
Now observe that when 158 turns the thickness 
of a tooth it turns 157 one tooth, and 157 turns 
156-a one tooth; and when 158 turns 12 teeth or 
one revolution, 156-a turns 12 teeth also, or i%6' 
or % of a revolution. Also, when 158 turns one 
tooth it turns 159 one tooth, and 159 turns 156-ft' 
one tooth; and when 158 turns 12 teeth or one 
revolution, 156-& also turns 12 teeth, or i%6 
% of a revolution. One revolution of 158 has 
turned 156-a and 156-& exactly the same, or ^ 
of a revolution, whether the motion was trans- 
mitted through a small gear or a large gear. Sa 
you see that the size of the go-between or in- 
termediary gear has nothing to do with the rate 
of motion transmitted to the next gear. The 
intermediary transmits to the follower a motion 
of as many teeth as it receives from the driver. 
Now let us suppose 156-a to be the driver trans-^ 
mitting motion through 157, 158 and 159 to 
156-&. When 156-a moves one tooth 156-& will' 
move one tooth; and when 156-a makes one rev- 
olution 156-& will make one revolution. Neither 
the number of intermediaries nor their size af- 
fects the relative speed of any gear in a train as 



^98 Poore's Foundation Course 

compared with that of any other. The relative 
speeds of any two gears in a train depend solely 
upon their relative number of teeth respectively. 
Again supposing 156-a to be the driver, when 
156-a makes one revolution 157 will make two be- 
cause it has half as many teeth as 156-a; 158 
will make one and a third revolutions, and 159 
will make two thirds of a revolution. While 
157 is making one revolution 159 makes one 
third of a revolution, and 157 turns three times 
while 159 turns once. 

e. We have seen that while 158 makes one 
revolution, 159 makes half a revolution because 

159 has 24 teeth while 158 has 12 teeth. We 
have seen also (a) that 159 and 160 are both 
tight gears on the same shaft, both turning with 
the shaft and both turning together; when 159 
makes a revolution 160 also makes a revolution. 

160 has 14 teeth and it meshes with 156-^ hav- 
ing 16 teeth. Now, when 158 makes one rev- 
olution, 159 makes half a revolution; 160 also 
makes half a revolution, and it moves seven 
teeth; it transmits a motion of seven teeth to 
156-^, or %6 of a revolution. 156-^ and 156-6 
have each 16 teeth; and we have seen that one 
revolution of 158 turns 156-6 i%6 of a revolu- 
tion, while 156-^ is turned %6 of a revolution. 
The difference is effected by the device of using 
two tight gears on the same shaft as an interme- 
diary, one of them receiving the motion and 
the other transmitting it to 156-^. It will be 
seen readily that by varying the relative sizes of 
these two tight gears acting as an intermediary. 



Poore's Foundation Course 99^ 

any desired change can be effected in the rela- 
tive speeds of 158 and 156-^. 

/. In transmitting power the intermediary 
gear transmits just what it receives, except of 
course a certain loss due to friction. // motion 
or speed is lost in transmitting, power is gained;: 
and if speed is gained power is lost. In a train 
of two gears, 158 and 159, 158 makes one revo- 
lution while 159 makes half a revolution. The 
speed transmitted to 159 is just half the speed 
of 158; but the power transmitted to 159 is 
double the power of 158. Add 156-& to the- 
train of gears making 159 an intermediary; the 
power transmitted to 156-& is just the same as if 
158 were directly in mesh with 156-& except for 
a certain loss by friction due to the intermediary.. 
The^ power transmitted is in an inverse (reversed) 
ratio to the motion or speed transmitted. We 
have seen that the speed transmitted to 156-& is 
^%6 of the speed of 158; the power transmit- 
ted to 156-b is therefore i%2 of the power of 
158. In other words, if 156-& were a tight gear 
on a shaft, its shaft would exert i^ times as. 
much force as is exerted upon the shaft of 158, 
but it would turn only ^ as fast as the shaft 
of 158. 

Sec. 80. Nos. 150, 151 and 152 show general', 
and detailed drawings of a frame garage with a 
bench across the width of the room under the 
window at the rear. This subject is introduced 
here not with the object of showing particular 
methods of construction. The purpose is rather 
to show how to indicate construction in draw- 
ing — any construction that it is desired to fol- 



100 Poore's Foundation Course 

low. The several views and sections show pret- 
ty clearly just what the construction is to be in 
this case ; and that is the whole purpose of mak- 
ing a drawing. For the purpose of this Course 
the foundation work is omitted from this study, 
having already included a drawing of a 
foundation. The drawings show a front eleva- 
tion, a rear elevation and a left hand side ele- 
vation in relation to the front elevation, and a sec- 
tional plan; a front elevation, plan and right 
hand side elevation of the window viewed from 
the outside; a section of sash muUion, sections 
•of bottom and side of window frame, two sec- 
tions of doors and door frame, section of sid- 
ing, section of a corner and two views of the 
bench construction. In the rear view and the 
left hand side view the siding and the roof are 
broken out that the frame and the bench in 
place may be shown. The encircled numbers on 
the drawing indicate parts of the construction, 
as follows: 

(i) Sills under sides, (2) sills under ends, 

(3) side plates, (4) end plates, (5) stud (six of 

them), (6) stud (four of them), (7) stud 

(fourteen of them), (8) rafters (twenty of 

them), (9) rafter peak board, (10) header over 

door, (11) corner boards (eight of them), (12) 

roof edge (all around), (13) roof boards, (20, 

21, 22 and 23) siding, (25) cap (two of them), 

(26) legs (four of them), {2^^ tie (two of 

them), (28) bolts (four of them), (29) bolts 

(two of them), (30) plank, (31) boards, (32) 

filling pieces (four of them), (34) end supports 

{two of them), (40) sash (upper and lower), 



Poore's Foundation Course 101 

(41) window sill, (42 and 43) window casing, 
(44 and 45) window jamb, (46, 47, 48 and 49) 
window stops, (50) right hand door, (51) left 
hand door, (52) door stop (top), (53) door 
jamb (top), (54) door casing (top), (55) 
weather strip ( top of door), (56) door casing, 
upright (two of them), (57) door jamb, sides 
(two of them), (58) door stops, sides (two of 
them), (59) finishing piece under door, (60) 
hinges hardware, (61) hasp and staple, (62) 
lock. 

a. Although these garage drawings are for 
a rather small and unimportant building, they 
present quite sufficient detail for the first subject 
in architectural drawing. Study the drawings 
carefully in connection with the foregoing ex- 
planation of the numbers until you are able to 
name and know how to place every piece of 
lumber in the building. Give very careful at- 
tention to the sections as a means of showing 
exactly how every construction is to be made; 
and note with care all details shown in all the 
drawings. Having studied the drawings until 
you understand every line in them, you are 
ready to reproduce them. Make a complete set 
of drawings of your own, and allow for them 
about twice as much space as they occupy on 
this sheet, if you draw them to the same scale. 
They are necessarily crowded together here. 
You can make your work present a better ap- 
pearance by giving it more space on the sheet. 
Pay very close attention to what may seem to 
you to be small things, and omit not the slightest 
detail. When you have mastered thoroughly 



102 Poore's Foundation Course 

this subject it will not be difficult for you to 
make drawings for a larger building. 

b. Having completed the drawings, the next 
step is to prepare what is called a "bill of ma- 
terial" for the building. It is simply an item- 
ized statement of all materials required and the 
quantity of each. The bill of material with a 
percentage added for labor cost forms the basis 
for your preliminary estimate of the cost of 
the building. The cost of materials is not the 
same in all parts of the country; and the per- 
centage to be added for labor cost varies even 
more. Go over the plans carefully and verify 
the following bill of material. Find the cost of 
the materials at the prevailing local prices, and 
add to that an estimate of the labor cost ac- 
cording to the best information you can get, for 
your estimate of the cost of the garage, not in- 
cluding the foundation. To find the cost of the 
foundation, you must first determine from the 
nature of the ground what will be necessary to 
make the foundation sufficiently secure. Then 
you can decide what material shall be used, and 
figure the number of cubic yards and the prob- 
able cost. The following is the bill of material 
(Sec. 25) :— 

ROUGH LUMBER— YELLOW PINE. 

(board measure) 

( I ) Sills 2 pes., 4x6x18= 72 ft. 

( 2) Sills 2 pes., 4 X 6 X 12 zn 48 ft. 

( 3) Plates 2 pes., 4x6x18= 72 ft. 

( 4) Plates 2 pes., 4x6x12= 48 ft. 

(5) Studs 6 pes., 6 X 6 X ID = 180 ft. 



Pooee's Foundation Course 103 

( 6) Studs 4 pes., 4 X 6 X lo = 8o f t. 

( 7) Studs 14 pes., 3 X 6 X 10 = 210 ft. 

( 8) Rafters 10 pes., 3x6x18 = 270 ft. 

( 9) Peak board i pe., i x 8 x 18 = 12 ft. 

(10) Header over 

door I pc. 4 X 6 X iQiiz 20 ft. 

(13) Roof boards, 357 sq.ft. plus waste 1= 380 ft. 
(32) Filling pes. bench, i pc., 2x4x8= 5 ft. 
(34) End suppports 

bench i pc, 2 x 6 x 8= 12 ft. 

Total 1409 ft. 

FINISHED LUMBER— YELLOW PINE. 
(20-21-22-23) Siding, exposed surface 650 
sq. ft., 20 per cent, allowance for lap 

and waste = 780 ft. 

(12) Roof edge 4 pes., 1x6x8'- 6"= 17 ft. 

(12) Roof edge 4 pes., 1x6x12" zz: 24 ft. 

(11) Comer boards . . .8 pes., 1x7x12 = 56 ft. 
(53) Door frame (top) i pc, 1x4x10 = 3 ft. 

(57) Door frame (sides ) 2 pes., I X4X 8 = 6 ft. 
(52) Door stop (top) . . I pc, IX3X 8 = 2 ft. 

(58) Door stop (sides) .2 pes., 1x3x8 = 4 ft. 
(54-56) Door casing . .3 pes., 1x7x10 = 18 ft. 
(55); Door weather strip, i pc, 1x3x10 = 2 it, 

(59) Door finishing pci pc, 1x3x10 =: 3 ft. 

(30) Bench plank i pc, 3x12x12 = 36 ft. 

(31) Bench boards. . .1x17 (total) x 12 = 17 ft. 

Total 968 ft. 

MILL WORK— YELLOW PINE. 
(50 Door, right hand, as per drawing. 
(51) Door, left hand, as per drawing. 



104 Poore's Foundation Course 

(151) Window frame and sash, as per drawing. 

(152) Two bench frames complete as per draw- 
ing, without Nos. 30-31-32. 

OTHER MATERIAL. I 

(60) Hinges — 4 prs. — 2 doz. lY^' — No. 14 i 

screws. 

(61) Hasp and staples. (62) Lock. | 
Nails — lod. (penny), 5 lbs.; 8d,, 25 lbs. 
Roofing paper — 4 rolls. Paint — 2^ gals. 



There have now been presented to you the 
facts, principles, rules and practices which con- 
stitute a basic knowledge of Mechanical Draw- 
ing and a general knowledge of the Arithmetic 
of Measures as applied to a wide range of shop 
problems. If you have followed with fidelity 
the suggestions contained in our foreword to 
the student, this knowledge is yours to a greater 
or to a less degree, according to your ability 
to grasp ideas readily and to apply them prac- 
tically. However it would be expecting too 
much to assume that you have mastered the 
Course, that you have exhausted the subject 
and have absorbed all that is taught herein by 
going over it just once. A second trip over 
the same route should prove very interesting 
and profitable, and a third for review will con- 
vince you that you have yet some work to do 
before you can feel sure that every element in 
this Foundation Course has been applied in your 
own foundation preparation for life's work. 



